ML Aggarwal Class 10 Maths Solutions Chapter 05 Quadratic Equations in One Variable

Question 1. In each of the following, determine whether the given numbers are roots of the given equations or not:
(i) \( x^2 - 5x + 6 = 0; 2, -3 \)
(ii) \( 3x^2 - 13x - 10 = 0; 5, -\frac{2}{3} \)
Answer:
(i) Substitute x = 2 into the left side of the equation to get:

\( \text{LHS} = 2^2 - 5 \times 2 + 6 = 4 - 10 + 6 = 0 = \text{RHS} \)

Therefore, 2 is a root. When x = -3:

\( \text{LHS} = (-3)^2 - 5 \times (-3) + 6 = 9 + 15 + 6 = 30 \neq \text{RHS} \)

So -3 is not a root. Hence, 2 is a root but -3 is not.

(ii) When x = 5:

\( \text{LHS} = 3 \times (5)^2 - 13 \times 5 - 10 = 75 - 65 - 10 = 0 = \text{RHS} \)

So 5 is a root. Substituting x = -\frac{2}{3}:

\( \text{LHS} = 3 \times \frac{4}{9} + 13 \times \frac{2}{3} - 10 = \frac{4}{3} + \frac{26}{3} - 10 = \frac{30}{3} - 10 = 0 = \text{RHS} \)

Therefore, both 5 and -\frac{2}{3} are roots.
In simple words: Replace x with each given number and calculate. If the result equals zero, that number is a root.

Exam Tip: Always substitute carefully and check your arithmetic twice - sign errors are common when dealing with negative numbers.

Question 2. In each of the following, determine whether the given numbers are solutions of the given equations or not:
(i) \( x^2 - 3\sqrt{3}x + 6 = 0; \sqrt{3}, -2\sqrt{3} \)
(ii) \( x^2 - \sqrt{2}x - 4 = 0; -\sqrt{2}, 2\sqrt{2} \)
Answer:
(i) Substituting x = \sqrt{3}:

\( \text{LHS} = (\sqrt{3})^2 - 3\sqrt{3} \times \sqrt{3} + 6 = 3 - 9 + 6 = 0 = \text{RHS} \)

So \sqrt{3} is a solution. For x = -2\sqrt{3}:

\( \text{LHS} = (-2\sqrt{3})^2 - 3\sqrt{3} \times (-2\sqrt{3}) + 6 = 12 + 18 + 6 = 36 \neq \text{RHS} \)

Therefore, \sqrt{3} is a solution but -2\sqrt{3} is not.

(ii) For x = -\sqrt{2}:

\( \text{LHS} = (-\sqrt{2})^2 - \sqrt{2} \times (-\sqrt{2}) - 4 = 2 + 2 - 4 = 0 = \text{RHS} \)

So -\sqrt{2} is a solution. When x = 2\sqrt{2}:

\( \text{LHS} = (2\sqrt{2})^2 - \sqrt{2} \times 2\sqrt{2} - 4 = 8 - 4 - 4 = 0 = \text{RHS} \)

Therefore, both -\sqrt{2} and 2\sqrt{2} are solutions.
In simple words: Plug in each value and simplify. When surds are involved, multiply carefully - the same surds multiply to give whole numbers.

Exam Tip: Pay close attention to operations with surds - remember that \( \sqrt{a} \times \sqrt{a} = a \) and handle signs with special care.

Question 3(i). If -\frac{1}{2} is the solution of the equation 3x^2 + 2kx - 3 = 0, find the value of k.
Answer: Since -\frac{1}{2} satisfies the equation, we substitute x = -\frac{1}{2}:

\( 3 \times \frac{1}{4} - k - 3 = 0 \)

\( \frac{3}{4} - k - 3 = 0 \)

\( k = \frac{3}{4} - 3 = \frac{3 - 12}{4} = -\frac{9}{4} \)

Therefore, k = -\frac{9}{4}.
In simple words: When a number is a solution, it must make the equation equal zero. Plug it in and solve for k.

Exam Tip: Always arrange terms carefully when solving for the unknown coefficient - a sign error will give the wrong value.

Question 3(ii). If \frac{2}{3} is the solution of the equation 7x^2 + kx - 3 = 0, find the value of k.
Answer: Since \frac{2}{3} satisfies the equation, substitute x = \frac{2}{3}:

\( 7 \times \frac{4}{9} + k \times \frac{2}{3} - 3 = 0 \)

\( \frac{28}{9} + \frac{2k}{3} - 3 = 0 \)

\( \frac{28}{9} - \frac{27}{9} + \frac{6k}{9} = 0 \)

\( 1 + 6k = 0 \)

\( k = -\frac{1}{6} \)

Therefore, k = -\frac{1}{6}.
In simple words: Replace x, find a common denominator, and solve for k by making the equation equal to zero.

Exam Tip: Convert all fractions to a common denominator before combining - this reduces arithmetic mistakes.

Question 4(i). If \sqrt{2} is a root of the equation kx^2 + \sqrt{2}x - 4 = 0, find the value of k.
Answer: Since \sqrt{2} is a root, it satisfies the equation:

\( k(\sqrt{2})^2 + \sqrt{2} \times \sqrt{2} - 4 = 0 \)

\( 2k + 2 - 4 = 0 \)

\( 2k - 2 = 0 \)

\( k = 1 \)

Therefore, k = 1.
In simple words: Put \sqrt{2} into the equation and simplify - remember that \sqrt{2} times \sqrt{2} gives 2.

Exam Tip: When working with surds, always multiply them out first before combining with other terms.

Question 4(ii). If a is the root of the equation x^2 - (a + b)x + k = 0, find the value of k.
Answer: Since a is a root, substitute x = a into the equation:

\( a^2 - (a + b)a + k = 0 \)

\( a^2 - a^2 - ab + k = 0 \)

\( k - ab = 0 \)

\( k = ab \)

Therefore, k = ab.
In simple words: Replace x with a and simplify by expanding the bracket. The a^2 terms cancel, leaving you with k = ab.

Exam Tip: Expand brackets carefully - the terms a^2 cancel here, which simplifies the problem significantly.

Question 5. If \frac{2}{3} and -3 are roots of the equation px^2 + 7x + q = 0, find the values of p and q.
Answer: Since \frac{2}{3} is a root, it satisfies the equation:

\( p \times \frac{4}{9} + 7 \times \frac{2}{3} + q = 0 \)

\( \frac{4p}{9} + \frac{14}{3} + q = 0 \)

\( \frac{4p + 42}{9} + q = 0 \) ... (i)

Since -3 is also a root:

\( p \times 9 + 7 \times (-3) + q = 0 \)

\( 9p - 21 + q = 0 \) ... (ii)

From equation (i): \( q = -\frac{4p + 42}{9} \)

Substitute into equation (ii):

\( 9p - 21 - \frac{4p + 42}{9} = 0 \)

\( \frac{81p - 189 - 4p - 42}{9} = 0 \)

\( 77p - 231 = 0 \)

\( p = 3 \)

Substituting p = 3 into equation (i):

\( q = -\frac{12 + 42}{9} = -\frac{54}{9} = -6 \)

Therefore, p = 3 and q = -6.
In simple words: Each root gives you one equation. Solve the two equations together to find both unknowns.

Exam Tip: Setting up two equations from two roots is the key strategy here - eliminate one variable and solve for the other first.

Exercise 5.2

Question 1(i). Solve the following equation by factorisation: x^2 - 3x - 10 = 0
Answer: We need to factor the quadratic. Look for two numbers that multiply to -10 and add to -3. These are -5 and 2:

\( x^2 - 5x + 2x - 10 = 0 \)

\( x(x - 5) + 2(x - 5) = 0 \)

\( (x + 2)(x - 5) = 0 \)

By the zero-product rule:

\( x + 2 = 0 \text{ or } x - 5 = 0 \)

\( x = -2 \text{ or } x = 5 \)

The roots are -2 and 5.
In simple words: Split the middle term into two parts, factor by grouping, then use the zero-product rule.

Exam Tip: Always check your factorisation by expanding back - if (x + 2)(x - 5) multiplies out correctly, you have the right factors.

Question 1(ii). Solve the following equation by factorisation: x(2x + 5) = 3
Answer: First, expand and rearrange to standard form:

\( 2x^2 + 5x = 3 \)

\( 2x^2 + 5x - 3 = 0 \)

Now factor by finding two numbers that multiply to -6 (which is 2 × -3) and add to 5. These are 6 and -1:

\( 2x^2 + 6x - x - 3 = 0 \)

\( 2x(x + 3) - 1(x + 3) = 0 \)

\( (2x - 1)(x + 3) = 0 \)

By the zero-product rule:

\( 2x - 1 = 0 \text{ or } x + 3 = 0 \)

\( x = \frac{1}{2} \text{ or } x = -3 \)

The roots are \frac{1}{2} and -3.
In simple words: Expand the brackets first, move everything to one side, then factor the resulting quadratic.

Exam Tip: When the leading coefficient is not 1, use the AC method - multiply the first and last coefficients to find the pair of numbers for splitting the middle term.

Question 2(i). Solve the following equation by factorisation: 3x^2 - 5x - 12 = 0
Answer: We need two numbers that multiply to -36 (which is 3 × -12) and add to -5. These are -9 and 4:

\( 3x^2 - 9x + 4x - 12 = 0 \)

\( 3x(x - 3) + 4(x - 3) = 0 \)

\( (x - 3)(3x + 4) = 0 \)

By the zero-product rule:

\( x - 3 = 0 \text{ or } 3x + 4 = 0 \)

\( x = 3 \text{ or } x = -\frac{4}{3} \)

The roots are 3 and -\frac{4}{3}.
In simple words: Find factors of ac = 3 × (-12) = -36 that also add to b = -5, then split and factor.

Exam Tip: The product ac gives you the target for finding the splitting pair - this technique works for all quadratics with any leading coefficient.

Question 2(ii). Solve the following equation by factorisation: 21x^2 - 8x - 4 = 0
Answer: We need two numbers that multiply to -84 (which is 21 × -4) and add to -8. These are -14 and 6:

\( 21x^2 - 14x + 6x - 4 = 0 \)

\( 7x(3x - 2) + 2(3x - 2) = 0 \)

\( (7x + 2)(3x - 2) = 0 \)

By the zero-product rule:

\( 7x + 2 = 0 \text{ or } 3x - 2 = 0 \)

\( x = -\frac{2}{7} \text{ or } x = \frac{2}{3} \)

The roots are -\frac{2}{7} and \frac{2}{3}.
In simple words: Use the same splitting method - find two numbers that give the correct product and sum, then factor by grouping.

Exam Tip: When you have fractional roots, always check by substituting back into the original equation to verify.

Question 3(i). Solve the following equation by factorisation: 3x^2 = x + 4
Answer: First, rearrange to standard form:

\( 3x^2 - x - 4 = 0 \)

We need two numbers that multiply to -12 (which is 3 × -4) and add to -1. These are -4 and 3:

\( 3x^2 - 4x + 3x - 4 = 0 \)

\( x(3x - 4) + 1(3x - 4) = 0 \)

\( (x + 1)(3x - 4) = 0 \)

By the zero-product rule:

\( x + 1 = 0 \text{ or } 3x - 4 = 0 \)

\( x = -1 \text{ or } x = \frac{4}{3} \)

The roots are -1 and \frac{4}{3}.
In simple words: Move all terms to one side first to get zero on the right, then factor and solve.

Exam Tip: Always check that you have written the equation in ax^2 + bx + c = 0 form before factoring - any other arrangement will confuse your work.

Question 3(ii). Solve the following equation by factorisation: x(6x - 1) = 35
Answer: Expand and rearrange to standard form:

\( 6x^2 - x = 35 \)

\( 6x^2 - x - 35 = 0 \)

We need two numbers that multiply to -210 (which is 6 × -35) and add to -1. These are -15 and 14:

\( 6x^2 - 15x + 14x - 35 = 0 \)

\( 3x(2x - 5) + 7(2x - 5) = 0 \)

\( (3x + 7)(2x - 5) = 0 \)

By the zero-product rule:

\( 3x + 7 = 0 \text{ or } 2x - 5 = 0 \)

\( x = -\frac{7}{3} \text{ or } x = \frac{5}{2} \)

The roots are -\frac{7}{3} and \frac{5}{2}.
In simple words: Expand the bracket, move everything to one side, then use the splitting method to factor.

Exam Tip: For larger values of ac, list factors systematically to find the pair that adds to b - this saves time compared to random guessing.

Question 4(i). Solve the following equation by factorisation: 6p^2 + 11p - 10 = 0
Answer: We need two numbers that multiply to -60 (which is 6 × -10) and add to 11. These are 15 and -4:

\( 6p^2 + 15p - 4p - 10 = 0 \)

\( 3p(2p + 5) - 2(2p + 5) = 0 \)

\( (2p + 5)(3p - 2) = 0 \)

By the zero-product rule:

\( 2p + 5 = 0 \text{ or } 3p - 2 = 0 \)

\( p = -\frac{5}{2} \text{ or } p = \frac{2}{3} \)

The roots are -\frac{5}{2} and \frac{2}{3}.
In simple words: The variable changes from x to p, but the factorisation process stays exactly the same.

Exam Tip: Remember that the method works for any variable name - the process of finding factors and grouping remains identical regardless of whether you use x, p, y, or another letter.

Question 4(ii). Solve the following equation by factorisation: \( \frac{2}{3}x^2 - \frac{1}{3}x = 1 \)
Answer: We are given \( \frac{2}{3}x^2 - \frac{1}{3}x = 1 \).

\( \Rightarrow \frac{2}{3}x^2 - \frac{1}{3}x - 1 = 0 \) (Writing as \( ax^2 + bx + c = 0 \))

\( \Rightarrow \frac{2}{3}x^2 \times 3 - \frac{1}{3}x \times 3 - 1 \times 3 = 0 \times 3 \) (Multiplying the equation by 3)

\( \Rightarrow 2x^2 - x - 3 = 0 \)

\( \Rightarrow 2x^2 - 3x + 2x - 3 = 0 \)

\( \Rightarrow x(2x - 3) + 1(2x - 3) = 0 \)

\( \Rightarrow (x + 1)(2x - 3) = 0 \) (Factorising left side)

\( \Rightarrow x + 1 = 0 \) or \( 2x - 3 = 0 \) (Zero-product rule)

\( \Rightarrow x = -1 \) or \( x = \frac{3}{2} \)
In simple words: Break the equation into two factors. Set each factor equal to zero and solve for x.

Exam Tip: Always multiply through by the denominator first to clear fractions, then proceed with standard factorisation steps.

Question 5(i). Solve the following equation by factorisation: \( 3(x - 2)^2 = 147 \)
Answer: We are given \( 3(x - 2)^2 = 147 \).

\( \Rightarrow 3(x^2 - 4x + 4) = 147 \)

\( \Rightarrow 3x^2 - 12x + 12 = 147 \)

\( \Rightarrow 3x^2 - 12x + 12 - 147 = 0 \) (Writing as \( ax^2 + bx + c = 0 \))

\( \Rightarrow 3x^2 - 12x - 135 = 0 \)

\( \Rightarrow \frac{3x^2 - 12x - 135}{3} = \frac{0}{3} \) (Dividing the complete equation by 3)

\( \Rightarrow x^2 - 4x - 45 = 0 \)

\( \Rightarrow x^2 - 9x + 5x - 45 = 0 \)

\( \Rightarrow x(x - 9) + 5(x - 9) = 0 \)

\( \Rightarrow (x - 9)(x + 5) = 0 \) (Factorising left side)

\( \Rightarrow x - 9 = 0 \) or \( x + 5 = 0 \) (Zero-product rule)

\( \Rightarrow x = 9 \) or \( x = -5 \)
In simple words: Expand the bracket, simplify to standard form, then factor and solve.

Exam Tip: After expanding, always collect all terms on one side and simplify by dividing out any common factor before factorising.

Question 5(ii). Solve the following equation by factorisation: \( \frac{1}{7}(3x - 5)^2 = 28 \)
Answer: We are given \( \frac{1}{7}(3x - 5)^2 = 28 \).

\( \Rightarrow \frac{1}{7}(9x^2 - 30x + 25) = 28 \)

\( \Rightarrow \frac{9}{7}x^2 - \frac{30}{7}x + \frac{25}{7} = 28 \)

\( \Rightarrow \frac{9}{7}x^2 \times 7 - \frac{30}{7}x \times 7 + \frac{25}{7} \times 7 = 28 \times 7 \) (Multiplying the complete equation by 7)

\( \Rightarrow 9x^2 - 30x + 25 = 196 \)

\( \Rightarrow 9x^2 - 30x + 25 - 196 = 0 \) (Writing as \( ax^2 + bx + c = 0 \))

\( \Rightarrow 9x^2 - 30x - 171 = 0 \)

\( \Rightarrow 9x^2 - 57x + 27x - 171 = 0 \)

\( \Rightarrow 3x(3x - 19) + 9(3x - 19) = 0 \)

\( \Rightarrow (3x + 9)(3x - 19) = 0 \) (Factorising left side)

\( \Rightarrow 3x + 9 = 0 \) or \( 3x - 19 = 0 \) (Zero - product rule)

\( \Rightarrow 3x = -9 \) or \( 3x = 19 \)

\( \Rightarrow x = -3 \) or \( x = \frac{19}{3} \)
In simple words: Clear the fraction by multiplying, expand the squared term, then solve using factorisation.

Exam Tip: Remember to multiply all terms by the denominator to eliminate fractions before proceeding with factorisation.

Question 6. Solve the following equation by factorisation: \( x^2 - 4x - 12 = 0 \) when \( x \in \mathbb{N} \)
Answer: We are given \( x^2 - 4x - 12 = 0 \).

\( \Rightarrow x^2 - 6x + 2x - 12 = 0 \)

\( \Rightarrow x(x - 6) + 2(x - 6) = 0 \)

\( \Rightarrow (x + 2)(x - 6) = 0 \) (Factorising left side)

\( \Rightarrow x + 2 = 0 \) or \( x - 6 = 0 \) (Zero-product rule)

\( \Rightarrow x = -2 \) or \( x = 6 \)

Since \( x \in \mathbb{N} \), \( x = -2 \) is not a root. Therefore, the root is \( x = 6 \).
In simple words: Factorise the expression, find both roots, then pick only the one that is a natural number.

Exam Tip: When a domain is given (like \( x \in \mathbb{N} \)), check both roots against that restriction and reject any that don't fit.

Question 7. Solve the following equation by factorisation: \( 2x^2 - 9x + 10 = 0 \), when
(i) \( x \in \mathbb{N} \)
(ii) \( x \in \mathbb{Q} \)
Answer: We are given \( 2x^2 - 9x + 10 = 0 \).

\( \Rightarrow 2x^2 - 5x - 4x + 10 = 0 \)

\( \Rightarrow x(2x - 5) - 2(2x - 5) = 0 \)

\( \Rightarrow (x - 2)(2x - 5) = 0 \) (Factorising left side)

\( \Rightarrow x - 2 = 0 \) or \( 2x - 5 = 0 \) (Zero-product rule)

\( \Rightarrow x = 2 \) or \( x = \frac{5}{2} \)

(i) Since \( x \in \mathbb{N} \), the root is \( x = 2 \).

(ii) Since \( x \in \mathbb{Q} \), both roots are valid: \( x = 2 \) or \( x = \frac{5}{2} \).
In simple words: Factorise to get two solutions. For natural numbers, pick only whole positive values. For rational numbers, include fractions too.

Exam Tip: Know the difference: \( \mathbb{N} \) means natural numbers (positive integers), while \( \mathbb{Q} \) includes all fractions and integers.

Question 8(i). Solve the following equation by factorisation: \( a^2x^2 + 2ax + 1 = 0 \), \( a \neq 0 \)
Answer: We are given \( a^2x^2 + 2ax + 1 = 0 \).

\( \Rightarrow a^2x^2 + ax + ax + 1 = 0 \)

\( \Rightarrow ax(ax + 1) + 1(ax + 1) = 0 \)

\( \Rightarrow (ax + 1)(ax + 1) = 0 \) (Factorising left side)

\( \Rightarrow ax + 1 = 0 \) (Zero-product rule)

\( \Rightarrow ax = -1 \)

\( \Rightarrow x = -\frac{1}{a} \)

The roots are \( x = -\frac{1}{a}, -\frac{1}{a} \) (a repeated root).
In simple words: This is a perfect square trinomial that factors into two identical terms, giving one repeated solution.

Exam Tip: When both factors are identical, you have a repeated root - write it with the correct multiplicity in your final answer.

Question 8(ii). Solve the following equation by factorisation: \( x^2 - (p + q)x + pq = 0 \)
Answer: We are given \( x^2 - (p + q)x + pq = 0 \).

\( \Rightarrow x^2 - px - qx + pq = 0 \)

\( \Rightarrow x(x - p) - q(x - p) = 0 \)

\( \Rightarrow (x - q)(x - p) = 0 \) (Factorising left side)

\( \Rightarrow x - q = 0 \) or \( x - p = 0 \) (Zero-product rule)

\( \Rightarrow x = q \) or \( x = p \)
In simple words: When a quadratic is written in the form \( x^2 - (\text{sum})x + (\text{product}) = 0 \), the roots are exactly the two values whose sum and product match.

Exam Tip: This is a standard pattern - if you see \( x^2 - (p+q)x + pq \), you can immediately write the factors as \( (x-p)(x-q) \).

Question 9. Solve the following equation by factorisation: \( a^2x^2 + (a^2 + b^2)x + b^2 = 0 \), \( a \neq 0 \)
Answer: We are given \( a^2x^2 + (a^2 + b^2)x + b^2 = 0 \).

\( \Rightarrow a^2x^2 + a^2x + b^2x + b^2 = 0 \)

\( \Rightarrow a^2x(x + 1) + b^2(x + 1) = 0 \)

\( \Rightarrow (a^2x + b^2)(x + 1) = 0 \) (Factorising left side)

\( \Rightarrow a^2x + b^2 = 0 \) or \( x + 1 = 0 \) (Zero-product rule)

\( \Rightarrow a^2x = -b^2 \) or \( x = -1 \)

\( \Rightarrow x = -\frac{b^2}{a^2} \) or \( x = -1 \)
In simple words: Split the middle term using the sum and product method, then factor by grouping into two binomials.

Exam Tip: When the middle term coefficient is a sum like \( a^2 + b^2 \), split it into its parts and group the first two terms separately from the last two.

Question 10(i). Solve the following equation by factorisation: \( \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \)
Answer: We are given \( \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \).

\( \Rightarrow \sqrt{3}x^2 + 7x + 3x + 7\sqrt{3} = 0 \)

\( \Rightarrow x(\sqrt{3}x + 7) + \sqrt{3}(\sqrt{3}x + 7) = 0 \)

\( \Rightarrow (x + \sqrt{3})(\sqrt{3}x + 7) = 0 \) (Factorising left side)

\( \Rightarrow x + \sqrt{3} = 0 \) or \( \sqrt{3}x + 7 = 0 \) (Zero - product rule)

\( \Rightarrow x = -\sqrt{3} \) or \( \sqrt{3}x = -7 \)

\( \Rightarrow x = -\sqrt{3} \) or \( x = -\frac{7}{\sqrt{3}} \)

\( \Rightarrow x = -\sqrt{3} \) or \( x = -\frac{7\sqrt{3}}{3} \)
In simple words: Factorise by splitting the middle term, then solve each factor equation separately.

Exam Tip: When rationalising a denominator with a square root, multiply both numerator and denominator by the same root.

Question 10(ii). Solve the following equation by factorisation: \( 4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0 \)
Answer: We are given \( 4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0 \).

\( \Rightarrow 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0 \)

\( \Rightarrow 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0 \)

\( \Rightarrow (\sqrt{3}x + 2)(4x - \sqrt{3}) = 0 \) (Factorising left side)

\( \Rightarrow \sqrt{3}x + 2 = 0 \) or \( 4x - \sqrt{3} = 0 \) (Zero-product rule)

\( \Rightarrow \sqrt{3}x = -2 \) or \( 4x = \sqrt{3} \)

\( \Rightarrow x = -\frac{2}{\sqrt{3}} \) or \( x = \frac{\sqrt{3}}{4} \)

\( \Rightarrow x = -\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \) or \( x = \frac{\sqrt{3}}{4} \)

\( \Rightarrow x = -\frac{2\sqrt{3}}{3} \) or \( x = \frac{\sqrt{3}}{4} \)
In simple words: Split the middle term carefully to identify the two factors, then apply the zero-product rule.

Exam Tip: When surds appear in the coefficient, rationalise denominators in the final answer by multiplying by the conjugate or the same surd.

Question 11(i). Solve the following equation by factorisation: \( x^2 - (1 + \sqrt{2})x + \sqrt{2} = 0 \)
Answer: We are given \( x^2 - (1 + \sqrt{2})x + \sqrt{2} = 0 \).

\( \Rightarrow x^2 - x - \sqrt{2}x + \sqrt{2} = 0 \)

\( \Rightarrow x(x - 1) - \sqrt{2}(x - 1) = 0 \)

\( \Rightarrow (x - \sqrt{2})(x - 1) = 0 \) (Factorising left side)

\( \Rightarrow x - \sqrt{2} = 0 \) or \( x - 1 = 0 \) (Zero-product rule)

\( \Rightarrow x = \sqrt{2} \) or \( x = 1 \)
In simple words: Split the middle term and factor by grouping, then set each factor to zero and solve.

Exam Tip: Notice the middle term is the negative of a sum: split it into its component parts and group terms with common factors.

Question 11(ii). Solve the following equation by factorisation: \( x + \frac{1}{x} = 2\frac{1}{20} \)
Answer: We are given \( x + \frac{1}{x} = 2\frac{1}{20} \).

Converting the mixed number: \( 2\frac{1}{20} = \frac{41}{20} \)

\( \Rightarrow x + \frac{1}{x} = \frac{41}{20} \)

\( \Rightarrow x \times x + \frac{1}{x} \times x = \frac{41}{20} \times x \)

\( \Rightarrow x^2 + 1 = \frac{41}{20}x \)

\( \Rightarrow 20(x^2 + 1) = 41x \)

\( \Rightarrow 20x^2 + 20 = 41x \)

\( \Rightarrow 20x^2 - 41x + 20 = 0 \) (Writing as \( ax^2 + bx + c = 0 \))

\( \Rightarrow 20x^2 - 25x - 16x + 20 = 0 \)

\( \Rightarrow 5x(4x - 5) - 4(4x - 5) = 0 \)

\( \Rightarrow (5x - 4)(4x - 5) = 0 \) (Factorising left side)

\( \Rightarrow 5x - 4 = 0 \) or \( 4x - 5 = 0 \) (Zero-product rule)

\( \Rightarrow 5x = 4 \) or \( 4x = 5 \)

\( \Rightarrow x = \frac{4}{5} \) or \( x = \frac{5}{4} \)
In simple words: Multiply through by x to clear the fraction, convert to standard form, then factorise and solve.

Exam Tip: When a variable appears in a denominator, multiply the entire equation by that variable (or its LCM) to eliminate the fraction before factorising.

Question 12(i). Solve the following equation by factorisation: \( \frac{2}{x^2} - \frac{5}{x} + 2 = 0 \), \( x \neq 0 \)
Answer: We are given \( \frac{2}{x^2} - \frac{5}{x} + 2 = 0 \).

\( \Rightarrow \frac{2 - 5x + 2x^2}{x^2} = 0 \)

\( \Rightarrow 2 - 5x + 2x^2 = 0 \times x^2 \)

\( \Rightarrow 2x^2 - 5x + 2 = 0 \)

\( \Rightarrow 2x^2 - 4x - x + 2 = 0 \)

\( \Rightarrow 2x(x - 2) - 1(x - 2) = 0 \)

\( \Rightarrow (2x - 1)(x - 2) = 0 \) (Factorising left side)

\( \Rightarrow 2x - 1 = 0 \) or \( x - 2 = 0 \) (Zero-product rule)

\( \Rightarrow 2x = 1 \) or \( x = 2 \)

\( \Rightarrow x = \frac{1}{2} \) or \( x = 2 \)
In simple words: Find a common denominator, clear it by multiplying, rearrange to standard form, then factor and solve.

Exam Tip: Always note the restriction (like \( x \neq 0 \)) and check that your final roots do not violate it.

Question 12(ii). Solve the following equation by factorisation: \( \frac{x^2}{15} - \frac{x}{3} - 10 = 0 \)
Answer: We are given \( \frac{x^2}{15} - \frac{x}{3} - 10 = 0 \).

\( \Rightarrow \frac{x^2 - x \times 5 - 10 \times 15}{15} = 0 \)

\( \Rightarrow x^2 - 5x - 150 = 0 \)

\( \Rightarrow x^2 - 15x + 10x - 150 = 0 \)

\( \Rightarrow x(x - 15) + 10(x - 15) = 0 \)

\( \Rightarrow (x + 10)(x - 15) = 0 \) (Factorising left side)

\( \Rightarrow x + 10 = 0 \) or \( x - 15 = 0 \) (Zero-product rule)

\( \Rightarrow x = -10 \) or \( x = 15 \)
In simple words: Get a common denominator (15), then multiply to remove fractions and solve using factorisation.

Exam Tip: When fractions have different denominators, always find the LCM to combine them into a single fraction before clearing the denominator.

Question 13(i). Solve the following equation by factorisation: \( 3x - \frac{8}{x} = 2 \)
Answer: We are given \( 3x - \frac{8}{x} = 2 \).

\( \Rightarrow \frac{3x^2 - 8}{x} = 2 \)

\( \Rightarrow 3x^2 - 8 = 2x \)

\( \Rightarrow 3x^2 - 2x - 8 = 0 \) (Writing as \( ax^2 + bx + c = 0 \))

\( \Rightarrow 3x^2 - 6x + 4x - 8 = 0 \)

\( \Rightarrow 3x(x - 2) + 4(x - 2) = 0 \)

\( \Rightarrow (3x + 4)(x - 2) = 0 \) (Factorising left side)

\( \Rightarrow 3x + 4 = 0 \) or \( x - 2 = 0 \) (Zero-product rule)

\( \Rightarrow 3x = -4 \) or \( x = 2 \)

\( \Rightarrow x = -\frac{4}{3} \) or \( x = 2 \)
In simple words: Multiply through by x to clear the fraction, rearrange to standard form, then factor and find both solutions.

Exam Tip: After multiplying to clear fractions, always move all terms to one side to get zero on the other before factorising.

Question 13(ii). Solve the following equation by factorisation: \( \frac{x + 2}{x + 3} = \frac{2x - 3}{3x - 7} \)
Answer: We are given \( \frac{x + 2}{x + 3} = \frac{2x - 3}{3x - 7} \).

\( \Rightarrow (x + 2) \times (3x - 7) = (2x - 3) \times (x + 3) \)

\( \Rightarrow 3x^2 - 7x + 6x - 14 = 2x^2 + 6x - 3x - 9 \)

\( \Rightarrow 3x^2 - x - 14 = 2x^2 + 3x - 9 \)

\( \Rightarrow 3x^2 - 2x^2 - x - 3x - 14 + 9 = 0 \)

\( \Rightarrow x^2 - 4x - 5 = 0 \)

\( \Rightarrow x^2 - 5x + x - 5 = 0 \)

\( \Rightarrow x(x - 5) + 1(x - 5) = 0 \)

\( \Rightarrow (x + 1)(x - 5) = 0 \)

\( \Rightarrow x + 1 = 0 \) or \( x - 5 = 0 \)

\( \Rightarrow x = -1 \) or \( x = 5 \)
In simple words: Cross-multiply to remove fractions, expand both sides, collect like terms, then solve by factorisation.

Exam Tip: When you have a rational equation (fractions equal to each other), cross-multiply first before expanding and simplifying.

Question 14(i). Solve the following equation by factorisation: \( \frac{3x - 9}{x + 3} - \frac{8}{2 - x} = 2 \)
Answer: We begin by finding a common denominator on the left side:
\( \Rightarrow \frac{8(2 - x) - 3(x + 3)}{(x + 3)(2 - x)} = 2 \)
\( \Rightarrow 8(2 - x) - 3(x + 3) = 2(x + 3)(2 - x) \)
\( \Rightarrow 16 - 8x - 3x - 9 = 2(2x - x^2 + 6 - 3x) \)
\( \Rightarrow 7 - 11x = 2(-x^2 - x + 6) \)
\( \Rightarrow 7 - 11x = -2x^2 - 2x + 12 \)
\( \Rightarrow 7 - 11x + 2x^2 + 2x - 12 = 0 \)
\( \Rightarrow 2x^2 - 9x - 5 = 0 \)
\( \Rightarrow 2x^2 - 10x + x - 5 = 0 \)
\( \Rightarrow 2x(x - 5) + 1(x - 5) = 0 \)
\( \Rightarrow (2x + 1)(x - 5) = 0 \)
\( \Rightarrow 2x + 1 = 0 \text{ or } x - 5 = 0 \)
\( \Rightarrow x = -\frac{1}{2} \text{ or } x = 5 \)
In simple words: Combine the fractions, cross-multiply, expand and simplify to get a quadratic equation. Then split and factor to find the two values of x.

Exam Tip: Always find a common denominator before cross-multiplying, and check that your roots do not make any denominator equal to zero.

Question 14(ii). Solve the following equation by factorisation: \( \frac{x}{x - 1} + \frac{x - 1}{x} = 2\frac{1}{2} \)
Answer: We express the mixed number as an improper fraction and find a common denominator:
\( \Rightarrow \frac{x \times x + (x - 1)(x - 1)}{x(x - 1)} = \frac{5}{2} \)
\( \Rightarrow x^2 + x^2 - x - x + 1 = \frac{5x(x - 1)}{2} \)
\( \Rightarrow 2x^2 - 2x + 1 = \frac{5x^2 - 5x}{2} \)
\( \Rightarrow 2(2x^2 - 2x + 1) = 5x^2 - 5x \)
\( \Rightarrow 4x^2 - 4x + 2 = 5x^2 - 5x \)
\( \Rightarrow 4x^2 - 5x^2 - 4x + 5x + 2 = 0 \)
\( \Rightarrow -x^2 + x + 2 = 0 \)
\( \Rightarrow x^2 - x - 2 = 0 \) (multiplying by - 1)
\( \Rightarrow x^2 - 2x + x - 2 = 0 \)
\( \Rightarrow x(x - 2) + 1(x - 2) = 0 \)
\( \Rightarrow (x + 1)(x - 2) = 0 \)
\( \Rightarrow x + 1 = 0 \text{ or } x - 2 = 0 \)
\( \Rightarrow x = -1 \text{ or } x = 2 \)
In simple words: Combine the two fractions by using a common denominator, then rearrange and factor to solve for x.

Exam Tip: When combining fractions with different denominators, multiply each numerator by the denominator of the other fraction to build your common denominator correctly.

Question 15(i). Solve the following equation by factorisation: \( \frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 3 \)
Answer: We combine the fractions on the left by finding a common denominator:
\( \Rightarrow \frac{(x + 1)(x + 2) + (x - 2)(x - 1)}{(x - 1)(x + 2)} = 3 \)
\( \Rightarrow \frac{x^2 + 2x + x + 2 + x^2 - x - 2x + 2}{(x - 1)(x + 2)} = 3 \)
\( \Rightarrow x^2 + 2x + x + 2 + x^2 - x - 2x + 2 = 3(x - 1)(x + 2) \)
\( \Rightarrow 2x^2 + 4 = 3(x^2 + 2x - x - 2) \)
\( \Rightarrow 2x^2 + 4 = 3(x^2 + x - 2) \)
\( \Rightarrow 2x^2 + 4 = 3x^2 + 3x - 6 \)
\( \Rightarrow 2x^2 - 3x^2 - 3x + 4 + 6 = 0 \)
\( \Rightarrow -x^2 - 3x + 10 = 0 \)
\( \Rightarrow x^2 + 3x - 10 = 0 \) (multiplying by - 1)
\( \Rightarrow x^2 + 5x - 2x - 10 = 0 \)
\( \Rightarrow x(x + 5) - 2(x + 5) = 0 \)
\( \Rightarrow (x - 2)(x + 5) = 0 \)
\( \Rightarrow x - 2 = 0 \text{ or } x + 5 = 0 \)
\( \Rightarrow x = 2 \text{ or } x = -5 \)
In simple words: Add the two fractions together, expand the numerator, then rearrange to form a quadratic and factor to find both roots.

Exam Tip: After combining fractions, always expand both the numerator (left side) and the product on the right side fully before simplifying, to avoid algebraic errors.

Question 15(ii). Solve the following equation by factorisation: \( \frac{1}{x - 3} - \frac{1}{x + 5} = \frac{1}{6} \)
Answer: We combine the fractions on the left side:
\( \Rightarrow \frac{(x + 5) - (x - 3)}{(x - 3)(x + 5)} = \frac{1}{6} \)
\( \Rightarrow \frac{x + 5 - x + 3}{(x - 3)(x + 5)} = \frac{1}{6} \)
\( \Rightarrow \frac{8}{(x - 3)(x + 5)} = \frac{1}{6} \)
\( \Rightarrow 8 \times 6 = (x - 3)(x + 5) \)
\( \Rightarrow 48 = x^2 + 5x - 3x - 15 \)
\( \Rightarrow 48 = x^2 + 2x - 15 \)
\( \Rightarrow x^2 + 2x - 15 - 48 = 0 \)
\( \Rightarrow x^2 + 2x - 63 = 0 \)
\( \Rightarrow x^2 + 9x - 7x - 63 = 0 \)
\( \Rightarrow x(x + 9) - 7(x + 9) = 0 \)
\( \Rightarrow (x - 7)(x + 9) = 0 \)
\( \Rightarrow x - 7 = 0 \text{ or } x + 9 = 0 \)
\( \Rightarrow x = 7 \text{ or } x = -9 \)
In simple words: Simplify the left side to get a single fraction, cross-multiply with the right side, then form and solve the resulting quadratic.

Exam Tip: When subtracting fractions, be careful with signs in the numerator - distribute the negative sign fully before simplifying.

Question 16(i). Solve the following equation by factorisation: \( \frac{a}{ax - 1} + \frac{b}{bx - 1} = a + b \), where \( a + b \neq 0 \), \( a \neq 0 \), \( ab \neq 0 \)
Answer: We rearrange by moving all terms to one side:
\( \Rightarrow \frac{a}{ax - 1} + \frac{b}{bx - 1} - a - b = 0 \)
\( \Rightarrow \left(\frac{a}{ax - 1} - b\right) + \left(\frac{b}{bx - 1} - a\right) = 0 \)
\( \Rightarrow \frac{a - b(ax - 1)}{ax - 1} + \frac{b - a(bx - 1)}{bx - 1} = 0 \)
\( \Rightarrow \frac{a - abx + b}{ax - 1} + \frac{b - abx + a}{bx - 1} = 0 \)
\( \Rightarrow (a - abx + b)\left(\frac{1}{ax - 1} + \frac{1}{bx - 1}\right) = 0 \)
\( \Rightarrow a - abx + b = 0 \text{ or } \frac{1}{ax - 1} + \frac{1}{bx - 1} = 0 \)
\( \Rightarrow a + b = abx \text{ or } \frac{bx - 1 + ax - 1}{(ax - 1)(bx - 1)} = 0 \)
\( \Rightarrow x = \frac{a + b}{ab} \text{ or } ax + bx - 2 = 0 \)
\( \Rightarrow x = \frac{a + b}{ab} \text{ or } x(a + b) = 2 \)
\( \Rightarrow x = \frac{a + b}{ab} \text{ or } x = \frac{2}{a + b} \)
In simple words: Isolate and factor out the common expression, then solve the two resulting equations separately to get both roots.

Exam Tip: When factoring out a common factor from an equation set equal to zero, remember that either the factor equals zero OR the remaining expression equals zero - both cases must be solved.

Question 16(ii). Solve the following equation by factorisation: \( \frac{1}{2a + b + 2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x} \)
Answer: We rearrange to move the \( \frac{1}{2x} \) term to the left:
\( \Rightarrow \frac{1}{2a + b + 2x} - \frac{1}{2x} = \frac{1}{2a} + \frac{1}{b} \)
\( \Rightarrow \frac{2x - (2a + b + 2x)}{(2a + b + 2x)(2x)} = \frac{b + 2a}{2ab} \)
\( \Rightarrow \frac{-(2a + b)}{(2a + b + 2x)(2x)} = \frac{b + 2a}{2ab} \)
\( \Rightarrow \frac{-1}{(2a + b + 2x)(2x)} = \frac{1}{2ab} \)
\( \Rightarrow -2ab = (2a + b + 2x)(2x) \)
\( \Rightarrow -2ab = 4ax + 2bx + 4x^2 \)
\( \Rightarrow -ab = 2ax + bx + 2x^2 \) (dividing by 2)
\( \Rightarrow 2x^2 + 2ax + bx + ab = 0 \)
\( \Rightarrow 2x(x + a) + b(x + a) = 0 \)
\( \Rightarrow (2x + b)(x + a) = 0 \)
\( \Rightarrow 2x + b = 0 \text{ or } x + a = 0 \)
\( \Rightarrow x = -\frac{b}{2} \text{ or } x = -a \)
In simple words: Move one term across, find a common denominator, cross-multiply, expand and then factor to find the two solutions.

Exam Tip: Keep track of negative signs carefully throughout the working, especially when cross-multiplying equations that have subtraction in the numerator.

Question 17. Solve the following equation by factorisation: \( \frac{1}{x + 6} + \frac{1}{x - 10} = \frac{3}{x - 4} \)
Answer: We combine the two fractions on the left:
\( \Rightarrow \frac{(x - 10) + (x + 6)}{(x + 6)(x - 10)} = \frac{3}{x - 4} \)
\( \Rightarrow \frac{2x - 4}{(x + 6)(x - 10)} = \frac{3}{x - 4} \)
\( \Rightarrow (2x - 4)(x - 4) = 3(x + 6)(x - 10) \)
\( \Rightarrow 2x^2 - 8x - 4x + 16 = 3(x^2 - 10x + 6x - 60) \)
\( \Rightarrow 2x^2 - 12x + 16 = 3(x^2 - 4x - 60) \)
\( \Rightarrow 2x^2 - 12x + 16 = 3x^2 - 12x - 180 \)
\( \Rightarrow 2x^2 - 3x^2 - 12x + 12x + 16 + 180 = 0 \)
\( \Rightarrow -x^2 + 196 = 0 \)
\( \Rightarrow x^2 = 196 \)
\( \Rightarrow x = \sqrt{196} = 14 \text{ or } x = -14 \)
In simple words: Add the left fractions, then cross-multiply and solve. The x terms cancel, leaving just the x - squared terms to find.

Exam Tip: When the x term cancels during simplification, you are left with a difference of squares or a simple quadratic in \( x^2 \) - solve it by taking the square root of both sides.

Question 18(i). Solve the following equation by factorisation: \( \sqrt{3x + 4} = x \)
Answer: We square both sides to remove the square root:
\( \Rightarrow 3x + 4 = x^2 \)
\( \Rightarrow x^2 - 3x - 4 = 0 \)
\( \Rightarrow x^2 - 4x + x - 4 = 0 \)
\( \Rightarrow x(x - 4) + 1(x - 4) = 0 \)
\( \Rightarrow (x + 1)(x - 4) = 0 \)
\( \Rightarrow x + 1 = 0 \text{ or } x - 4 = 0 \)
\( \Rightarrow x = -1 \text{ or } x = 4 \)

Since we squared both sides, we must check both roots in the original equation \( \sqrt{3x + 4} = x \):

Checking for \( x = -1 \):
\( \Rightarrow \sqrt{3(-1) + 4} = -1 \)
\( \Rightarrow \sqrt{-3 + 4} = -1 \)
\( \Rightarrow \sqrt{1} = -1 \)
\( \Rightarrow 1 = -1 \) (This is false)

Checking for \( x = 4 \):
\( \Rightarrow \sqrt{3(4) + 4} = 4 \)
\( \Rightarrow \sqrt{12 + 4} = 4 \)
\( \Rightarrow \sqrt{16} = 4 \)
\( \Rightarrow 4 = 4 \) (This is true)

Since \( x = -1 \) does not satisfy the original equation, it is rejected.
In simple words: Square both sides to clear the square root. Solve the quadratic by factoring. Always check your answers back in the original equation because squaring can introduce extra solutions.

Exam Tip: After squaring an equation with a square root, checking both roots in the original (not the squared) equation is mandatory - one or both roots may be extraneous and must be rejected if they don't work.

Question 18(ii). Solve the following equation by factorisation: \( \sqrt{x(x - 7)} = 3\sqrt{2} \)
Answer: We square both sides:
\( \Rightarrow x(x - 7) = 18 \)
\( \Rightarrow x^2 - 7x = 18 \)
\( \Rightarrow x^2 - 7x - 18 = 0 \)
\( \Rightarrow x^2 - 9x + 2x - 18 = 0 \)
\( \Rightarrow x(x - 9) + 2(x - 9) = 0 \)
\( \Rightarrow (x + 2)(x - 9) = 0 \)
\( \Rightarrow x + 2 = 0 \text{ or } x - 9 = 0 \)
\( \Rightarrow x = -2 \text{ or } x = 9 \)

We check both roots in the original equation \( \sqrt{x(x - 7)} = 3\sqrt{2} \):

Checking for \( x = -2 \):
\( \Rightarrow \sqrt{-2(-2 - 7)} = 3\sqrt{2} \)
\( \Rightarrow \sqrt{-2(-9)} = 3\sqrt{2} \)
\( \Rightarrow \sqrt{18} = 3\sqrt{2} \)
\( \Rightarrow 3\sqrt{2} = 3\sqrt{2} \) (This is true)

Checking for \( x = 9 \):
\( \Rightarrow \sqrt{9(9 - 7)} = 3\sqrt{2} \)
\( \Rightarrow \sqrt{9 \times 2} = 3\sqrt{2} \)
\( \Rightarrow \sqrt{18} = 3\sqrt{2} \)
\( \Rightarrow 3\sqrt{2} = 3\sqrt{2} \) (This is true)

Both roots satisfy the original equation.
In simple words: Square both sides, solve the quadratic by factoring, and check both roots in the original equation to confirm they are valid.

Exam Tip: Simplify surds in the original equation first - recognising that \( (3\sqrt{2})^2 = 9 \times 2 = 18 \) makes the working clearer and faster.

Question 19. Use the substitution \( y = 3x + 1 \) to solve for x: \( 5(3x + 1)^2 + 6(3x + 1) - 8 = 0 \)
Answer: We substitute \( y = 3x + 1 \) into the equation:
\( \Rightarrow 5y^2 + 6y - 8 = 0 \)
\( \Rightarrow 5y^2 + 10y - 4y - 8 = 0 \)
\( \Rightarrow 5y(y + 2) - 4(y + 2) = 0 \)
\( \Rightarrow (5y - 4)(y + 2) = 0 \)
\( \Rightarrow 5y - 4 = 0 \text{ or } y + 2 = 0 \)
\( \Rightarrow y = \frac{4}{5} \text{ or } y = -2 \)

Now we substitute back \( y = 3x + 1 \):

When \( y = \frac{4}{5} \):
\( \Rightarrow 3x + 1 = \frac{4}{5} \)
\( \Rightarrow 3x = \frac{4}{5} - 1 \)
\( \Rightarrow 3x = -\frac{1}{5} \)
\( \Rightarrow x = -\frac{1}{15} \)

When \( y = -2 \):
\( \Rightarrow 3x + 1 = -2 \)
\( \Rightarrow 3x = -3 \)
\( \Rightarrow x = -1 \)
In simple words: Replace the repeated expression with a single variable, solve the resulting simpler quadratic, then substitute back to find x.

Exam Tip: Substitution is most useful when a single expression appears multiple times (especially when raised to different powers) - always substitute back at the end to recover the original variable.

Question 20. Find the values of x if \( p + 1 = 0 \) and \( x^2 + px - 6 = 0 \)
Answer: From the first condition, we find the value of p:
\( \Rightarrow p + 1 = 0 \)
\( \Rightarrow p = -1 \)

Now we substitute \( p = -1 \) into the second equation:
\( \Rightarrow x^2 + (-1)x - 6 = 0 \)
\( \Rightarrow x^2 - x - 6 = 0 \)
\( \Rightarrow x^2 - 3x + 2x - 6 = 0 \)
\( \Rightarrow x(x - 3) + 2(x - 3) = 0 \)
\( \Rightarrow (x + 2)(x - 3) = 0 \)
\( \Rightarrow x + 2 = 0 \text{ or } x - 3 = 0 \)
\( \Rightarrow x = -2 \text{ or } x = 3 \)
In simple words: Use the first condition to find p, then substitute that value into the second equation and solve the resulting quadratic.

Exam Tip: Always solve the simpler equation (here, \( p + 1 = 0 \)) first before substituting into the more complex one - this reduces the chance of error.

Question 21. Find the values of x if p + 7 = 0, q - 12 = 0 and x² + px + q = 0.
Answer: Since p + 7 = 0 and q - 12 = 0, we get p = -7 and q = 12. Substituting these values into x² + px + q = 0:
\( x^2 + (-7)x + 12 = 0 \)
\( \Rightarrow x^2 - 7x + 12 = 0 \)
\( \Rightarrow x^2 - 4x - 3x + 12 = 0 \)
\( \Rightarrow x(x - 4) - 3(x - 4) = 0 \)
\( \Rightarrow (x - 3)(x - 4) = 0 \) (Factorising left side)
\( \Rightarrow x - 3 = 0 \) or \( x - 4 = 0 \) (Zero-product rule)
\( \Rightarrow x = 3 \) or \( x = 4 \)
In simple words: Plug in the values p = -7 and q = 12. Then factor the resulting quadratic to find x = 3 and x = 4.

Exam Tip: Always substitute the given expressions for p and q first, then solve the quadratic using factorisation or the formula method.

Question 22. If x = p is a solution of the equation x(2x + 5) = 3, then find the values of p.
Answer: Since x = p satisfies the equation x(2x + 5) = 3, we substitute x = p:
\( p(2p + 5) = 3 \)
\( \Rightarrow 2p^2 + 5p = 3 \)
\( \Rightarrow 2p^2 + 5p - 3 = 0 \)
\( \Rightarrow 2p^2 + 6p - p - 3 = 0 \)
\( \Rightarrow 2p(p + 3) - 1(p + 3) = 0 \)
\( \Rightarrow (2p - 1)(p + 3) = 0 \) (Factorising left side)
\( \Rightarrow 2p - 1 = 0 \) or \( p + 3 = 0 \) (Zero-product rule)
\( \Rightarrow 2p = 1 \) or \( p = -3 \)
\( \Rightarrow p = \frac{1}{2} \) or \( p = -3 \)
In simple words: Replace x with p in the given equation, then solve the quadratic to get p = 1/2 or p = -3.

Exam Tip: Check both solutions by substituting back into the original equation to confirm they satisfy it.

Question 23. If x = 3 is a solution of the equation (k + 2)x² - kx + 6 = 0, find the value of k. Hence, find the other root of the equation.
Answer: Since x = 3 satisfies the equation (k + 2)x² - kx + 6 = 0, we substitute x = 3:
\( (k + 2)(3)^2 - k(3) + 6 = 0 \)
\( \Rightarrow 9(k + 2) - 3k + 6 = 0 \)
\( \Rightarrow 9k + 18 - 3k + 6 = 0 \)
\( \Rightarrow 6k + 24 = 0 \)
\( \Rightarrow 6k = -24 \)
\( \Rightarrow k = -4 \)
Now substitute k = -4 into the original equation to find the other root:
\( (-4 + 2)x^2 - (-4)x + 6 = 0 \)
\( \Rightarrow -2x^2 + 4x + 6 = 0 \)
\( \Rightarrow 2x^2 - 4x - 6 = 0 \) (Multiplying equation by -1)
\( \Rightarrow 2x^2 - 6x + 2x - 6 = 0 \)
\( \Rightarrow 2x(x - 3) + 2(x - 3) = 0 \)
\( \Rightarrow (2x + 2)(x - 3) = 0 \) (Factorising left side)
\( \Rightarrow 2x + 2 = 0 \) or \( x - 3 = 0 \) (Zero-product rule)
\( \Rightarrow 2x = -2 \) or \( x = 3 \)
\( \Rightarrow x = -1 \) or \( x = 3 \)
In simple words: First, substitute x = 3 to find k = -4. Then use this value to solve the full equation and get the other root, x = -1.

Exam Tip: Remember that one root is already given, so focus on finding k first, then solve for the remaining root using factorisation or the quadratic formula.

Exercise 5.3

Question 1(i). Solve the following equations by using formula: 2x² - 7x + 6 = 0
Answer: The given equation is 2x² - 7x + 6 = 0. Comparing it with ax² + bx + c = 0, we get a = 2, b = -7, c = 6. Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 2 \times 6}}{2 \times 2} \)
\( \Rightarrow x = \frac{7 \pm \sqrt{49 - 48}}{4} \)
\( \Rightarrow x = \frac{7 \pm \sqrt{1}}{4} \)
\( \Rightarrow x = \frac{7 + 1}{4} \) or \( x = \frac{7 - 1}{4} \)
\( \Rightarrow x = \frac{8}{4} \) or \( x = \frac{6}{4} \)
\( \Rightarrow x = 2 \) or \( x = \frac{3}{2} \)
In simple words: Use the quadratic formula by identifying a, b, and c from the equation. Calculate the discriminant, then apply the formula to get the two roots.

Exam Tip: Always check that the discriminant (b² - 4ac) is positive before taking its square root. Simplify fractions fully in the final answer.

Question 1(ii). Solve the following equations by using formula: 2x² - 6x + 3 = 0
Answer: The given equation is 2x² - 6x + 3 = 0. Comparing it with ax² + bx + c = 0, we get a = 2, b = -6, c = 3. Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 2 \times 3}}{2 \times 2} \)
\( \Rightarrow x = \frac{6 \pm \sqrt{36 - 24}}{4} \)
\( \Rightarrow x = \frac{6 \pm \sqrt{12}}{4} \)
\( \Rightarrow x = \frac{6 + \sqrt{12}}{4} \) or \( x = \frac{6 - \sqrt{12}}{4} \)
\( \Rightarrow x = \frac{6 + 2\sqrt{3}}{4} \) or \( x = \frac{6 - 2\sqrt{3}}{4} \)
\( \Rightarrow x = \frac{3 + \sqrt{3}}{2} \) or \( x = \frac{3 - \sqrt{3}}{2} \)
In simple words: Apply the quadratic formula after finding a = 2, b = -6, c = 3. Simplify the square root and reduce the fraction to get the final answers.

Exam Tip: When simplifying surds like √12, break them into perfect square factors (√12 = √4 × √3 = 2√3) to make further simplification easier.

Question 2(i). Solve the following equations by using formula: 256x² - 32x + 1 = 0
Answer: The given equation is 256x² - 32x + 1 = 0. Comparing it with ax² + bx + c = 0, we get a = 256, b = -32, c = 1. Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-(-32) \pm \sqrt{(-32)^2 - 4 \times 256 \times 1}}{2 \times 256} \)
\( \Rightarrow x = \frac{32 \pm \sqrt{1024 - 1024}}{512} \)
\( \Rightarrow x = \frac{32 \pm \sqrt{0}}{512} \)
\( \Rightarrow x = \frac{32}{512} \) or \( x = \frac{32}{512} \)
\( \Rightarrow x = \frac{1}{16} \) or \( x = \frac{1}{16} \)
In simple words: When the discriminant equals zero, both roots become identical. This means the equation has one repeated root.

Exam Tip: A discriminant of zero indicates a perfect square trinomial and a repeated root. Always reduce fractions to lowest terms.

Question 2(ii). Solve the following equations by using formula: 25x² + 30x + 7 = 0
Answer: The given equation is 25x² + 30x + 7 = 0. Comparing it with ax² + bx + c = 0, we get a = 25, b = 30, c = 7. Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-(30) \pm \sqrt{(30)^2 - 4 \times 25 \times 7}}{2 \times 25} \)
\( \Rightarrow x = \frac{-30 \pm \sqrt{900 - 700}}{50} \)
\( \Rightarrow x = \frac{-30 \pm \sqrt{200}}{50} \)
\( \Rightarrow x = \frac{-30 + \sqrt{200}}{50} \) or \( x = \frac{-30 - \sqrt{200}}{50} \)
\( \Rightarrow x = \frac{-30 + 10\sqrt{2}}{50} \) or \( x = \frac{-30 - 10\sqrt{2}}{50} \)
\( \Rightarrow x = \frac{-3 + \sqrt{2}}{5} \) or \( x = \frac{-3 - \sqrt{2}}{5} \)
In simple words: Identify the coefficients, compute the discriminant, and apply the formula. Simplify surds and fractions in the final result.

Exam Tip: Factor out common numbers from the numerator to simplify (e.g., 10√2 can be factored out of both -30 ± 10√2).

Question 3(i). Solve the following equations by using formula: 2x² + √5x - 5 = 0
Answer: The given equation is 2x² + √5x - 5 = 0. Comparing it with ax² + bx + c = 0, we get a = 2, b = √5, c = -5. Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-\sqrt{5} \pm \sqrt{(\sqrt{5})^2 - 4 \times 2 \times (-5)}}{2 \times 2} \)
\( \Rightarrow x = \frac{-\sqrt{5} \pm \sqrt{5 + 40}}{4} \)
\( \Rightarrow x = \frac{-\sqrt{5} \pm \sqrt{45}}{4} \)
\( \Rightarrow x = \frac{-\sqrt{5} + \sqrt{45}}{4} \) or \( x = \frac{-\sqrt{5} - \sqrt{45}}{4} \)
\( \Rightarrow x = \frac{-\sqrt{5} + 3\sqrt{5}}{4} \) or \( x = \frac{-\sqrt{5} - 3\sqrt{5}}{4} \)
\( \Rightarrow x = \frac{2\sqrt{5}}{4} \) or \( x = \frac{-4\sqrt{5}}{4} \)
\( \Rightarrow x = \frac{\sqrt{5}}{2} \) or \( x = -\sqrt{5} \)
In simple words: When coefficients involve surds, apply the quadratic formula carefully. Simplify the square root in the discriminant and combine like terms.

Exam Tip: Always simplify surds in the discriminant first (√45 = 3√5), then combine the terms in the numerator before final reduction.

Question 3(ii). Solve the following equations by using formula: √3x² + 10x - 8√3 = 0
Answer: The given equation is √3x² + 10x - 8√3 = 0. Comparing it with ax² + bx + c = 0, we get a = √3, b = 10, c = -8√3. Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-(10) \pm \sqrt{(10)^2 - 4 \times \sqrt{3} \times (-8\sqrt{3})}}{2 \times \sqrt{3}} \)
\( \Rightarrow x = \frac{-10 \pm \sqrt{100 + 96}}{2\sqrt{3}} \)
\( \Rightarrow x = \frac{-10 \pm \sqrt{196}}{2\sqrt{3}} \)
\( \Rightarrow x = \frac{-10 + 14}{2\sqrt{3}} \) or \( x = \frac{-10 - 14}{2\sqrt{3}} \)
\( \Rightarrow x = \frac{4}{2\sqrt{3}} \) or \( x = \frac{-24}{2\sqrt{3}} \)
\( \Rightarrow x = \frac{2}{\sqrt{3}} \) or \( x = \frac{-12}{\sqrt{3}} \)
\( \Rightarrow x = \frac{2\sqrt{3}}{3} \) or \( x = \frac{-12\sqrt{3}}{3} \)
\( \Rightarrow x = \frac{2\sqrt{3}}{3} \) or \( x = -4\sqrt{3} \)
In simple words: Calculate the discriminant carefully when the equation contains surds. Simplify fully, including rationalising the denominator if needed.

Exam Tip: When rationalising denominators with surds, multiply both numerator and denominator by the surd to get a rational denominator.

Question 4(i). Solve the following equations by using formula: \( \frac{x - 2}{x + 2} + \frac{x + 2}{x - 2} = 4 \)
Answer: Given: \( \frac{x - 2}{x + 2} + \frac{x + 2}{x - 2} = 4 \)
\( \Rightarrow \frac{(x - 2)^2 + (x + 2)^2}{(x - 2)(x + 2)} = 4 \)
\( \Rightarrow \frac{x^2 + 4 - 4x + x^2 + 4 + 4x}{x^2 - 2x + 2x - 4} = 4 \)
\( \Rightarrow \frac{2x^2 + 8}{x^2 - 4} = 4 \)
\( \Rightarrow 2x^2 + 8 = 4(x^2 - 4) \)
\( \Rightarrow 2x^2 + 8 = 4x^2 - 16 \)
\( \Rightarrow 2x^2 - 4x^2 + 8 + 16 = 0 \)
\( \Rightarrow -2x^2 + 24 = 0 \)
\( \Rightarrow 2x^2 - 24 = 0 \) (Multiplying equation by -1)
The equation is 2x² - 24 = 0. Comparing it with ax² + bx + c = 0, we get a = 2, b = 0, c = -24. Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-0 \pm \sqrt{0^2 - 4 \times 2 \times (-24)}}{2 \times 2} \)
\( \Rightarrow x = \frac{\pm \sqrt{192}}{4} \)
\( \Rightarrow x = \frac{\pm 8\sqrt{3}}{4} \)
\( \Rightarrow x = 2\sqrt{3} \) or \( x = -2\sqrt{3} \)
In simple words: Add the fractions by finding a common denominator, simplify, and then solve the resulting quadratic equation.

Exam Tip: When working with rational equations, always find a common denominator first, then cross-multiply to create a standard quadratic equation before applying the formula.

Question 4(ii). Solve the following equations by using formula: \( \frac{x + 1}{x + 3} = \frac{3x + 2}{2x + 3} \)
Answer: Given: \( \frac{x + 1}{x + 3} = \frac{3x + 2}{2x + 3} \)
\( \Rightarrow (x + 1)(2x + 3) = (3x + 2)(x + 3) \) (On cross multiplication)
\( \Rightarrow 2x^2 + 3x + 2x + 3 = 3x^2 + 9x + 2x + 6 \)
\( \Rightarrow 2x^2 + 5x + 3 = 3x^2 + 11x + 6 \)
\( \Rightarrow 2x^2 - 3x^2 + 5x - 11x + 3 - 6 = 0 \)
\( \Rightarrow -x^2 - 6x - 3 = 0 \)
\( \Rightarrow x^2 + 6x + 3 = 0 \) (Multiplying equation by -1)
The equation is x² + 6x + 3 = 0. Comparing it with ax² + bx + c = 0, we get a = 1, b = 6, c = 3. Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 3}}{2 \times 1} \)
\( \Rightarrow x = \frac{-6 \pm \sqrt{36 - 12}}{2} \)
\( \Rightarrow x = \frac{-6 \pm \sqrt{24}}{2} \)
\( \Rightarrow x = \frac{-6 + \sqrt{24}}{2} \) or \( x = \frac{-6 - \sqrt{24}}{2} \)
\( \Rightarrow x = \frac{-6 + 2\sqrt{6}}{2} \) or \( x = \frac{-6 - 2\sqrt{6}}{2} \)
\( \Rightarrow x = -3 + \sqrt{6} \) or \( x = -3 - \sqrt{6} \)
In simple words: Cross-multiply the two fractions to form a standard quadratic, then solve using the formula by first computing the discriminant.

Exam Tip: Ensure all like terms are combined carefully. Simplify surds in the final answer, and always check that the solutions don't make any denominator zero.

Question 5(i). Solve the following equations by using formula: a(x² + 1) = (a² + 1)x, a ≠ 0
Answer: Given: a(x² + 1) = (a² + 1)x. First, convert the equation to the form ax² + bx + c = 0:
\( \Rightarrow ax^2 + a = a^2x + x \)
\( \Rightarrow ax^2 - a^2x - x + a = 0 \)
\( \Rightarrow ax^2 - (a^2 + 1)x + a = 0 \)
Comparing it with ax² + bx + c = 0, we get a = a, b = -(a² + 1), c = a. Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-[-(a^2 + 1)] \pm \sqrt{[-(a^2 + 1)]^2 - 4 \times a \times a}}{2 \times a} \)
\( \Rightarrow x = \frac{a^2 + 1 \pm \sqrt{(a^2 + 1)^2 - 4a^2}}{2a} \)
\( \Rightarrow x = \frac{a^2 + 1 \pm \sqrt{a^4 + 1 + 2a^2 - 4a^2}}{2a} \)
\( \Rightarrow x = \frac{a^2 + 1 \pm \sqrt{a^4 + 1 - 2a^2}}{2a} \)
\( \Rightarrow x = \frac{a^2 + 1 \pm \sqrt{(a^2 - 1)^2}}{2a} \)
\( \Rightarrow x = \frac{a^2 + 1 \pm (a^2 - 1)}{2a} \)
\( \Rightarrow x = \frac{a^2 + 1 + a^2 - 1}{2a} \) or \( x = \frac{a^2 + 1 - a^2 + 1}{2a} \)
\( \Rightarrow x = \frac{2a^2}{2a} \) or \( x = \frac{2}{2a} \)
\( \Rightarrow x = a \) or \( x = \frac{1}{a} \)
In simple words: Rearrange the given equation into standard form, then apply the quadratic formula. Simplify the discriminant, which turns out to be a perfect square.

Exam Tip: Notice that the discriminant reduces to a perfect square (a² - 1)², making the solution process simpler. Always check if surds can be simplified to perfect squares.

Question 5(ii). Solve the following equations by using formula: 4x² - 4ax + (a² - b²) = 0
Answer: We have the equation 4x² - 4ax + (a² - b²) = 0.

Matching it with ax² + bx + c = 0, we get a = 4, b = -4a, c = a² - b².

Applying the formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies x = \frac{-(-4a) \pm \sqrt{(-4a)^2 - 4 \times 4 \times (a^2 - b^2)}}{2 \times 4} \)

\( \implies x = \frac{4a \pm \sqrt{16a^2 - 16(a^2 - b^2)}}{8} \)

\( \implies x = \frac{4a \pm \sqrt{16a^2 - 16a^2 + 16b^2}}{8} \)

\( \implies x = \frac{4a \pm \sqrt{16b^2}}{8} \)

\( \implies x = \frac{4a \pm 4b}{8} \)

\( \implies x = \frac{4a + 4b}{8} \text{ or } x = \frac{4a - 4b}{8} \)

Therefore, the roots are \( x = \frac{a + b}{2}, \frac{a - b}{2} \).
In simple words: We matched the given equation with the standard form and used the quadratic formula. After simplifying the discriminant and the entire expression, we got two roots that depend on the values of a and b.

Exam Tip: Always match coefficients carefully with the standard form ax² + bx + c = 0 before applying the formula, and simplify radical expressions fully before stating the final roots.

Question 6(i). Solve the following equations by using formula: x - 1/x = 3, x ≠ 0
Answer: Given: \( x - \frac{1}{x} = 3, x \neq 0 \)

Multiplying both sides by x (taking L.C.M):

\( \implies x^2 - 1 = 3x \)

\( \implies x^2 - 3x - 1 = 0 \)

Matching with ax² + bx + c = 0, we get a = 1, b = -3, c = -1.

Using the formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times (-1)}}{2 \times 1} \)

\( \implies x = \frac{3 \pm \sqrt{9 + 4}}{2} \)

\( \implies x = \frac{3 \pm \sqrt{13}}{2} \)

Therefore, the roots are \( x = \frac{3 + \sqrt{13}}{2}, \frac{3 - \sqrt{13}}{2} \).
In simple words: We first cleared the fraction by multiplying throughout by x, then rearranged to get a quadratic. The quadratic formula gave us two roots involving \( \sqrt{13} \).

Exam Tip: Always check for restrictions (x ≠ 0 here) before clearing denominators, and verify that your final roots do not violate these restrictions.

Question 6(ii). Solve the following equations by using formula: 1/x + 1/(x - 2) = 3, x ≠ 0, 2
Answer: Given: \( \frac{1}{x} + \frac{1}{x - 2} = 3, x \neq 0, 2 \)

Taking L.C.M on the left side:

\( \implies \frac{x - 2 + x}{x(x - 2)} = 3 \)

\( \implies \frac{2x - 2}{x^2 - 2x} = 3 \)

\( \implies 2x - 2 = 3(x^2 - 2x) \)

\( \implies 2x - 2 = 3x^2 - 6x \)

\( \implies 3x^2 - 6x - 2x + 2 = 0 \)

\( \implies 3x^2 - 8x + 2 = 0 \)

Matching with ax² + bx + c = 0, we get a = 3, b = -8, c = 2.

Using the formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 3 \times 2}}{2 \times 3} \)

\( \implies x = \frac{8 \pm \sqrt{64 - 24}}{6} \)

\( \implies x = \frac{8 \pm \sqrt{40}}{6} \)

\( \implies x = \frac{8 \pm 2\sqrt{10}}{6} \)

Therefore, the roots are \( x = \frac{4 + \sqrt{10}}{3}, \frac{4 - \sqrt{10}}{3} \).
In simple words: We found a common denominator for the two fractions, then simplified to get a quadratic equation. Using the formula, we found the two solutions involving \( \sqrt{10} \).

Exam Tip: When combining fractions, always clearly state the L.C.M and check that your restrictions (x ≠ 0, 2) are preserved in the final answer.

Question 7. Solve for x: 2(2x - 1)/(x + 3) - 3(x + 3)/(2x - 1) = 5, x ≠ -3, 1/2
Answer: Given: \( 2\left(\frac{2x - 1}{x + 3}\right) - 3\left(\frac{x + 3}{2x - 1}\right) = 5 \)

Let \( y = \frac{2x - 1}{x + 3} \). Then the equation becomes:

\( \implies 2y - \frac{3}{y} = 5 \)

\( \implies 2y^2 - 3 = 5y \)

\( \implies 2y^2 - 5y - 3 = 0 \)

Matching with ay² + by + c = 0, we get a = 2, b = -5, c = -3.

Using the formula:

\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 2 \times (-3)}}{2 \times 2} \)

\( \implies y = \frac{5 \pm \sqrt{25 + 24}}{4} \)

\( \implies y = \frac{5 \pm \sqrt{49}}{4} \)

\( \implies y = \frac{5 \pm 7}{4} \)

\( \implies y = 3 \text{ or } y = -\frac{1}{2} \)

Since \( y = \frac{2x - 1}{x + 3} \):

When \( y = 3 \): \( 3 = \frac{2x - 1}{x + 3} \)
\( \implies 3(x + 3) = 2x - 1 \)
\( \implies 3x + 9 = 2x - 1 \)
\( \implies x = -10 \)

When \( y = -\frac{1}{2} \): \( -\frac{1}{2} = \frac{2x - 1}{x + 3} \)
\( \implies -(x + 3) = 2(2x - 1) \)
\( \implies -x - 3 = 4x - 2 \)
\( \implies -5x = 1 \)
\( \implies x = -\frac{1}{5} \)

Therefore, \( x = -10 \text{ or } x = -\frac{1}{5} \).
In simple words: We used substitution to simplify the equation by letting y equal one of the fractions. After solving the resulting quadratic, we substituted back to find the original variable x.

Exam Tip: Substitution is a powerful technique for equations with repeated fractional expressions; always remember to substitute back to find x from y before stating the final answer.

Question 8. Solve the following quadratic equations for x and give your answer correct to 2 decimal places: (i) x² - 5x - 10 = 0 (ii) x² + 7x = 7
Answer:
(i) The given equation is x² - 5x - 10 = 0.

Matching with ax² + bx + c = 0, we get a = 1, b = -5, c = -10.

Using the formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 1 \times (-10)}}{2 \times 1} \)

\( \implies x = \frac{5 \pm \sqrt{25 + 40}}{2} \)

\( \implies x = \frac{5 \pm \sqrt{65}}{2} \)

Since \( \sqrt{65} \approx 8.062 \):

\( \implies x = \frac{5 + 8.062}{2} \text{ or } x = \frac{5 - 8.062}{2} \)

\( \implies x = \frac{13.062}{2} \text{ or } x = \frac{-3.062}{2} \)

\( \implies x = 6.531 \text{ or } x = -1.531 \)

\( \implies x = 6.53 \text{ or } x = -1.53 \) (correct to 2 decimal places)

(ii) Given: x² + 7x = 7
\( \implies x^2 + 7x - 7 = 0 \)

Matching with ax² + bx + c = 0, we get a = 1, b = 7, c = -7.

Using the formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies x = \frac{-(7) \pm \sqrt{(7)^2 - 4 \times 1 \times (-7)}}{2 \times 1} \)

\( \implies x = \frac{-7 \pm \sqrt{49 + 28}}{2} \)

\( \implies x = \frac{-7 \pm \sqrt{77}}{2} \)

Since \( \sqrt{77} \approx 8.775 \):

\( \implies x = \frac{-7 + 8.775}{2} \text{ or } x = \frac{-7 - 8.775}{2} \)

\( \implies x = \frac{1.775}{2} \text{ or } x = \frac{-15.775}{2} \)

\( \implies x = 0.888 \text{ or } x = -7.888 \)

\( \implies x = 0.89 \text{ or } x = -7.89 \) (correct to 2 decimal places)
In simple words: For both parts, we identified the coefficients, applied the quadratic formula, computed the square root using tables, and rounded the final answers to exactly 2 decimal places as instructed.

Exam Tip: Always use tables or a calculator for square root values when decimals are required, and round the final answer to the specified number of decimal places—never round intermediate steps.

Question 9. Solve the following equations by using quadratic formula and give answer correct to 2 decimal places: (i) 4x² - 5x - 3 = 0 (ii) x² - 7x + 3 = 0
Answer:
(i) The given equation is 4x² - 5x - 3 = 0.

Matching with ax² + bx + c = 0, we get a = 4, b = -5, c = -3.

Using the formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 4 \times (-3)}}{2 \times 4} \)

\( \implies x = \frac{5 \pm \sqrt{25 + 48}}{8} \)

\( \implies x = \frac{5 \pm \sqrt{73}}{8} \)

Since \( \sqrt{73} \approx 8.54 \):

\( \implies x = \frac{5 + 8.54}{8} \text{ or } x = \frac{5 - 8.54}{8} \)

\( \implies x = \frac{13.54}{8} \text{ or } x = \frac{-3.54}{8} \)

\( \implies x = 1.69 \text{ or } x = -0.44 \) (correct to 2 decimal places)

(ii) Given: x² - 7x + 3 = 0

Matching with ax² + bx + c = 0, we get a = 1, b = -7, c = 3.

Using the formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 1 \times 3}}{2 \times 1} \)

\( \implies x = \frac{7 \pm \sqrt{49 - 12}}{2} \)

\( \implies x = \frac{7 \pm \sqrt{37}}{2} \)

Since \( \sqrt{37} \approx 6.08 \):

\( \implies x = \frac{7 + 6.08}{2} \text{ or } x = \frac{7 - 6.08}{2} \)

\( \implies x = \frac{13.08}{2} \text{ or } x = \frac{0.92}{2} \)

\( \implies x = 6.54 \text{ or } x = 0.46 \) (correct to 2 decimal places)
In simple words: For both equations, we matched them with the standard form, applied the formula, looked up square root values, and rounded to 2 decimal places.

Exam Tip: When rounding to 2 decimal places, compute at least one extra decimal place before rounding to ensure accuracy; always show the unrounded value before stating the rounded final answer.

Question 10. Solve the following quadratic equations and give your answer correct to two significant figures: (i) x² - 4x - 8 = 0 (ii) x - 18/x = 6
Answer:
(i) The given equation is x² - 4x - 8 = 0.

Matching with ax² + bx + c = 0, we get a = 1, b = -4, c = -8.

Using the formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times (-8)}}{2 \times 1} \)

\( \implies x = \frac{4 \pm \sqrt{16 + 32}}{2} \)

\( \implies x = \frac{4 \pm \sqrt{48}}{2} \)

Since \( \sqrt{48} \approx 6.928 \):

\( \implies x = \frac{4 + 6.928}{2} \text{ or } x = \frac{4 - 6.928}{2} \)

\( \implies x = \frac{10.928}{2} \text{ or } x = \frac{-2.928}{2} \)

\( \implies x = 5.464 \text{ or } x = -1.464 \)

\( \implies x = 5.5 \text{ or } x = -1.5 \) (correct to 2 significant figures)

(ii) Given: \( x - \frac{18}{x} = 6 \)

Multiplying by x:
\( \implies x^2 - 18 = 6x \)
\( \implies x^2 - 6x - 18 = 0 \)

Matching with ax² + bx + c = 0, we get a = 1, b = -6, c = -18.

Using the formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times (-18)}}{2 \times 1} \)

\( \implies x = \frac{6 \pm \sqrt{36 + 72}}{2} \)

\( \implies x = \frac{6 \pm \sqrt{108}}{2} \)

Since \( \sqrt{108} \approx 10.392 \):

\( \implies x = \frac{6 + 10.392}{2} \text{ or } x = \frac{6 - 10.392}{2} \)

\( \implies x = \frac{16.392}{2} \text{ or } x = \frac{-4.392}{2} \)

\( \implies x = 8.196 \text{ or } x = -2.195 \)

\( \implies x = 8.2 \text{ or } x = -2.2 \) (correct to 2 significant figures)
In simple words: We found the coefficients, applied the formula, computed square roots, and rounded each final value to exactly 2 significant figures.

Exam Tip: Significant figures and decimal places are different—for significant figures, count from the first non-zero digit; always verify your rounding by checking how many non-zero digits appear in the rounded answer.

Question 11. Solve the equation 2x² - 10x + 5 = 0 and give your answer correct to 3 significant figures.
Answer: Matching the equation 2x² - 10x + 5 = 0 with ax² + bx + c = 0, we get a = 2, b = -10, c = 5.

Using the formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \times 2 \times 5}}{2 \times 2} \)

\( = \frac{10 \pm \sqrt{100 - 40}}{4} \)

\( = \frac{10 \pm \sqrt{60}}{4} \)

\( = \frac{10 \pm 2\sqrt{15}}{4} \)

\( = \frac{2(5 \pm \sqrt{15})}{4} \)

\( = \frac{5 \pm \sqrt{15}}{2} \)

Since \( \sqrt{15} \approx 3.87 \):

\( = \frac{5 \pm 3.87}{2} \)

\( \implies x = \frac{5 + 3.87}{2} \text{ or } x = \frac{5 - 3.87}{2} \)

\( \implies x = \frac{8.87}{2} \text{ or } x = \frac{1.13}{2} \)

\( \implies x = 4.435 \text{ or } x = 0.565 \)

\( \implies x = 4.44 \text{ or } x = 0.565 \) (correct to 3 significant figures)
In simple words: We substituted the coefficients into the formula, simplified the expression under the square root, and rounded each solution to 3 significant figures.

Exam Tip: When simplifying radicals, factor out perfect squares first (e.g., \( \sqrt{60} = 2\sqrt{15} \)); this helps reduce computational error and makes rounding more accurate at the end.

Exercise 5.4

Question 1. Find the discriminant of the following quadratic equations and hence find the nature of roots :
(i) \( 3x^2 - 5x - 2 = 0 \)
(ii) \( 2x^2 - 3x + 5 = 0 \)
(iii) \( 16x^2 - 40x + 25 = 0 \)
(iv) \( 2x^2 + 15x + 30 = 0 \)
Answer:
(i) Start with \( 3x^2 - 5x - 2 = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 3 \), \( b = -5 \), \( c = -2 \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (-5)^2 - 4 \times 3 \times (-2) \)

\( = 25 + 24 \)

\( = 49 > 0 \)

Discriminant = 49; hence, the given equation has two distinct real roots.

(ii) Start with \( 2x^2 - 3x + 5 = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 2 \), \( b = -3 \), \( c = 5 \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (-3)^2 - 4 \times 2 \times 5 \)

\( = 9 - 40 \)

\( = -31 < 0 \)

Discriminant = -31; hence, the given equation has no real roots.

(iii) Start with \( 16x^2 - 40x + 25 = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 16 \), \( b = -40 \), \( c = 25 \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (-40)^2 - 4 \times 16 \times 25 \)

\( = 1600 - 1600 \)

\( = 0 \)

Discriminant = 0; hence, the given equation has two equal real roots.

(iv) Start with \( 2x^2 + 15x + 30 = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 2 \), \( b = 15 \), \( c = 30 \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (15)^2 - 4 \times 2 \times 30 \)

\( = 225 - 240 \)

\( = -15 < 0 \)

Discriminant = -15; hence, the given equation has no real roots.
In simple words: The discriminant tells you what kind of roots an equation has. If it is positive, there are two different roots. If it is zero, both roots are the same. If it is negative, there are no real roots at all.

Exam Tip: Always compute the discriminant first before finding roots - it saves time and helps you identify the nature of solutions immediately.

Question 2. Discuss the nature of the roots of the following quadratic equations :
(i) \( 3x^2 - 4\sqrt{3}x + 4 = 0 \)
(ii) \( x^2 - \frac{1}{2}x + 4 = 0 \)
(iii) \( -2x^2 + x + 1 = 0 \)
(iv) \( 2\sqrt{3}x^2 - 5x + \sqrt{3} = 0 \)
Answer:
(i) Start with \( 3x^2 - 4\sqrt{3}x + 4 = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 3 \), \( b = -4\sqrt{3} \), \( c = 4 \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (-4\sqrt{3})^2 - 4 \times 3 \times 4 \)

\( = 48 - 48 \)

\( = 0 \)

Hence the given equation has two equal real roots.

(ii) Start with \( x^2 - \frac{1}{2}x + 4 = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 1 \), \( b = -\frac{1}{2} \), \( c = 4 \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = \left(-\frac{1}{2}\right)^2 - 4 \times 1 \times 4 \)

\( = \frac{1}{4} - 16 \)

\( = \frac{1 - 64}{4} \) (Taking L.C.M.)

\( = -\frac{63}{4} < 0 \)

Hence the given equation has no real roots.

(iii) Start with \( -2x^2 + x + 1 = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = -2 \), \( b = 1 \), \( c = 1 \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (1)^2 - 4 \times (-2) \times 1 \)

\( = 1 + 8 \)

\( = 9 > 0 \)

Hence, the given equation has two distinct real roots.

(iv) Start with \( 2\sqrt{3}x^2 - 5x + \sqrt{3} = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 2\sqrt{3} \), \( b = -5 \), \( c = \sqrt{3} \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (-5)^2 - 4 \times 2\sqrt{3} \times \sqrt{3} \)

\( = 25 - 24 \)

\( = 1 > 0 \)

Hence, the given equation has two distinct real roots.
In simple words: Use the formula \( b^2 - 4ac \) to find the discriminant. A positive value means two different roots exist. Zero means the roots are identical. A negative value means no real roots exist.

Exam Tip: When the equation has irrational coefficients (like \( \sqrt{3} \)), simplify carefully - products of identical square roots (e.g., \( \sqrt{3} \times \sqrt{3} = 3 \)) often emerge in the calculation.

Question 3. Find the nature of roots of the following quadratic equations :
(i) \( x^2 - \frac{1}{2}x - \frac{1}{2} = 0 \)
(ii) \( x^2 - 2\sqrt{3}x - 1 = 0 \)

If real roots exist, find them.
Answer:
(i) Start with \( x^2 - \frac{1}{2}x - \frac{1}{2} = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 1 \), \( b = -\frac{1}{2} \), \( c = -\frac{1}{2} \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = \left(-\frac{1}{2}\right)^2 - 4 \times 1 \times \left(-\frac{1}{2}\right) \)

\( = \frac{1}{4} + 2 \)

\( = \frac{1 + 8}{4} \)

\( = \frac{9}{4} \)

Since Discriminant > 0, hence the given equation has two distinct real roots. The roots are given by:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies x = \frac{-\left(-\frac{1}{2}\right) \pm \sqrt{\left(-\frac{1}{2}\right)^2 - 4 \times 1 \times \left(-\frac{1}{2}\right)}}{2 \times 1} \)

\( \implies x = \frac{\frac{1}{2} \pm \sqrt{\frac{1}{4} + 2}}{2} \)

\( \implies x = \frac{\frac{1}{2} \pm \sqrt{\frac{9}{4}}}{2} \)

\( \implies x = \frac{\frac{1}{2} \pm \frac{3}{2}}{2} \)

\( \implies x = \frac{\frac{1 + 3}{2}}{2} \) or \( \frac{\frac{1 - 3}{2}}{2} \)

\( \implies x = \frac{\frac{4}{2}}{2} \) or \( \frac{\frac{-2}{2}}{2} \)

\( \implies x = \frac{4}{4} \) or \( -\frac{2}{4} \)

\( \implies x = 1 \) or \( -\frac{1}{2} \)

Hence roots of the given equations are \( 1, -\frac{1}{2} \).

(ii) Start with \( x^2 - 2\sqrt{3}x - 1 = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 1 \), \( b = -2\sqrt{3} \), \( c = -1 \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (-2\sqrt{3})^2 - 4 \times 1 \times (-1) \)

\( = 12 + 4 \)

\( = 16 \)

Since Discriminant > 0, hence given equation have real and distinct roots. The roots are given by:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

\( \implies x = \frac{-(-2\sqrt{3}) \pm \sqrt{(-2\sqrt{3})^2 - 4 \times 1 \times (-1)}}{2 \times 1} \)

\( \implies x = \frac{2\sqrt{3} \pm \sqrt{12 + 4}}{2} \)

\( \implies x = \frac{2\sqrt{3} \pm \sqrt{16}}{2} \)

\( \implies x = \frac{2\sqrt{3} + 4}{2} \) or \( \frac{2\sqrt{3} - 4}{2} \)

\( \implies x = \sqrt{3} + 2 \) or \( \sqrt{3} - 2 \)

Hence roots of the given equations are \( \sqrt{3} + 2, \sqrt{3} - 2 \).
In simple words: Once you know real roots exist (discriminant is positive), use the quadratic formula with care. Simplify the square root first, then carefully divide each term in the numerator by 2.

Exam Tip: When applying the quadratic formula, ensure the discriminant is fully simplified before taking its square root - this prevents arithmetic errors in the final answer.

Question 4. Without solving the following quadratic equations, find the value of 'p' for which the given equations have real and equal roots :
(i) \( px^2 - 4x + 3 = 0 \)
(ii) \( x^2 + (p - 3)x + p = 0 \)
Answer:
(i) Start with \( px^2 - 4x + 3 = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = p \), \( b = -4 \), \( c = 3 \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (-4)^2 - 4 \times p \times 3 \)

\( = 16 - 12p \)

For equal roots, discriminant = 0

\( \implies 16 - 12p = 0 \)

\( \implies 16 = 12p \)

\( \implies p = \frac{16}{12} \)

\( \implies p = \frac{4}{3} \)

Hence the value of p is \( \frac{4}{3} \).

(ii) Start with \( x^2 + (p - 3)x + p = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 1 \), \( b = (p - 3) \), \( c = p \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (p - 3)^2 - 4 \times 1 \times p \)

\( = p^2 + 9 - 6p - 4p \)

\( = p^2 + 9 - 10p \)

For equal roots, discriminant = 0

\( \implies p^2 + 9 - 10p = 0 \)

\( \implies p^2 - 10p + 9 = 0 \)

\( \implies p^2 - 9p - p + 9 = 0 \)

\( \implies p(p - 9) - 1(p - 9) = 0 \)

\( \implies (p - 1)(p - 9) = 0 \)

\( \implies (p - 1) = 0 \) or \( p - 9 = 0 \)

\( \implies p = 1 \) or \( p = 9 \)

Hence the value of p is 1, 9.
In simple words: For equal roots, the discriminant must always be zero. Set \( b^2 - 4ac = 0 \) and solve for the unknown parameter to find all values that give equal roots.

Exam Tip: After finding parameter values, verify that the coefficient of \( x^2 \) is not zero - if it is, the equation is no longer quadratic and the equal roots condition may not apply.

Question 5. Find the values of k for which each of the following quadratic equation has equal roots :
(i) \( x^2 + 4kx + (k^2 - k + 2) = 0 \)
(ii) \( (k - 4)x^2 + 2(k - 4)x + 4 = 0 \)
Answer:
(i) Start with \( x^2 + 4kx + (k^2 - k + 2) = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 1 \), \( b = 4k \), \( c = (k^2 - k + 2) \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (4k)^2 - 4 \times 1 \times (k^2 - k + 2) \)

\( = 16k^2 - 4(k^2 - k + 2) \)

\( = 16k^2 - 4k^2 + 4k - 8 \)

\( = 12k^2 + 4k - 8 \)

For equal roots, discriminant = 0

\( \implies 12k^2 + 4k - 8 = 0 \)

\( \implies 12k^2 + 12k - 8k - 8 = 0 \)

\( \implies 12k(k + 1) - 8(k + 1) = 0 \)

\( \implies (k + 1)(12k - 8) = 0 \)

\( \implies k + 1 = 0 \) or \( 12k - 8 = 0 \)

\( \implies k = -1 \) or \( k = \frac{8}{12} \)

\( \implies k = -1 \) or \( k = \frac{2}{3} \)

Hence, the value of k is \( -1, \frac{2}{3} \).

(ii) Start with \( (k - 4)x^2 + 2(k - 4)x + 4 = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = k - 4 \), \( b = 2(k - 4) \), \( c = 4 \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (2k - 8)^2 - 4 \times (k - 4) \times 4 \)

\( = 4k^2 + 64 - 32k - 16(k - 4) \)

\( = 4k^2 - 32k - 16k + 64 + 64 \)

\( = 4k^2 - 48k + 128 \)

For equal roots, discriminant = 0

\( \implies 4k^2 - 48k + 128 = 0 \)

\( \implies 4(k^2 - 12k + 32) = 0 \)

\( \implies k^2 - 12k + 32 = 0 \)

\( \implies k^2 - 8k - 4k + 32 = 0 \)

\( \implies k(k - 8) - 4(k - 8) = 0 \)

\( \implies (k - 8)(k - 4) = 0 \)

\( \implies k = 4 \) or \( k = 8 \)

Since \( k \neq 4 \), (as that would make \( a = (k - 4) = 0 \) and roots would become infinite), \( k = 8 \) is the only valid solution.

Hence, the value of k is 8.
In simple words: Find when the discriminant equals zero. Factor the resulting quadratic and solve for k. Always check that the leading coefficient is not zero - if the parameter equals that value, it is not a valid solution.

Exam Tip: When the coefficient of \( x^2 \) contains the parameter, always verify that the parameter values found do not make this coefficient zero - such values must be rejected.

Question 6. Find the value(s) of m for which each of the following quadratic equation has real and equal roots :
(i) \( (3m + 1)x^2 + 2(m + 1)x + m = 0 \)
(ii) \( x^2 + 2(m - 1)x + (m + 5) = 0 \)
Answer:
(i) Start with \( (3m + 1)x^2 + 2(m + 1)x + m = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 3m + 1 \), \( b = 2(m + 1) \), \( c = m \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (2m + 2)^2 - 4 \times (3m + 1) \times m \)

\( = 4m^2 + 4 + 8m - 4m(3m + 1) \)

\( = 4m^2 + 4 + 8m - 12m^2 - 4m \)

\( = 4m^2 - 12m^2 + 8m - 4m + 4 \)

\( = -8m^2 + 4m + 4 \)

For equal roots, discriminant = 0

\( \implies -8m^2 + 4m + 4 = 0 \)

\( \implies -4(2m^2 - m - 1) = 0 \)

\( \implies 2m^2 - m - 1 = 0 \)

\( \implies 2m^2 - 2m + m - 1 = 0 \)

\( \implies 2m(m - 1) + 1(m - 1) = 0 \)

\( \implies (2m + 1)(m - 1) = 0 \)

\( \implies 2m + 1 = 0 \) or \( m - 1 = 0 \)

\( \implies m = -\frac{1}{2} \) or \( m = 1 \)

Hence, the value of m is \( -\frac{1}{2} \) and 1.

(ii) Start with \( x^2 + 2(m - 1)x + (m + 5) = 0 \). Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 1 \), \( b = 2(m - 1) \), \( c = m + 5 \).

\( \implies \) Discriminant \( = b^2 - 4ac \)

\( = (2m - 2)^2 - 4 \times 1 \times (m + 5) \)

\( = 4m^2 + 4 - 8m - 4(m + 5) \)

\( = 4m^2 + 4 - 8m - 4m - 20 \)

\( = 4m^2 - 12m - 16 \)

For equal roots, discriminant = 0

\( \implies 4m^2 - 12m - 16 = 0 \)

\( \implies 4(m^2 - 3m - 4) = 0 \)

\( \implies m^2 - 3m - 4 = 0 \)

\( \implies m^2 - 4m + m - 4 = 0 \)

\( \implies m(m - 4) + 1(m - 4) = 0 \)

\( \implies (m - 4)(m + 1) = 0 \)

\( \implies m - 4 = 0 \) or \( m + 1 = 0 \)

\( \implies m = 4 \) or \( m = -1 \)

Hence, the value of m is 4, -1.
In simple words: Set the discriminant equal to zero and simplify the resulting quadratic equation in the parameter. Factor and solve to find all parameter values that create equal roots.

Exam Tip: After computing the discriminant, factor out common numerical factors before setting it to zero - this simplifies the algebra and reduces calculation mistakes.

Question 7. Find the values of k for which each of the following quadratic equation has equal roots :

Note: The source document does not provide sub-parts or an answer for Question 7. This question heading appears incomplete in the fragment provided.

Question 7. Find the value(s) of k for which the quadratic equation 9x² + kx + 1 = 0 has equal roots. Also find the roots for those values of k in each case.
Answer: For the equation 9x² + kx + 1 = 0, we identify a = 9, b = k, c = 1. Setting the discriminant equal to zero: \( b^2 - 4ac = 0 \)
\( k^2 - 36 = 0 \)
\( (k + 6)(k - 6) = 0 \)
This gives k = 6 or k = -6.

When k = 6, the equation becomes 9x² + 6x + 1 = 0. Using the quadratic formula:
\( x = \frac{-6 \pm \sqrt{36 - 36}}{18} = \frac{-6}{18} = -\frac{1}{3} \)
Both roots equal \( -\frac{1}{3} \).

When k = -6, the equation becomes 9x² - 6x + 1 = 0. Using the quadratic formula:
\( x = \frac{6 \pm \sqrt{36 - 36}}{18} = \frac{6}{18} = \frac{1}{3} \)
Both roots equal \( \frac{1}{3} \).
In simple words: When the discriminant equals zero, a quadratic has two identical roots. For this equation, k must be 6 or -6. When k = 6, both roots are \( -\frac{1}{3} \); when k = -6, both roots are \( \frac{1}{3} \).

Exam Tip: Set discriminant to zero for equal roots. Always verify by substituting your k value back into the quadratic formula.

Question 7(ii). Find the value(s) of k for which the quadratic equation x² - 2kx + 7k - 12 = 0 has equal roots. Also find the roots for those values of k in each case.
Answer: For the equation x² - 2kx + 7k - 12 = 0, we have a = 1, b = -2k, c = 7k - 12. Setting discriminant = 0:
\( (-2k)^2 - 4(1)(7k - 12) = 0 \)
\( 4k^2 - 28k + 48 = 0 \)
\( k^2 - 7k + 12 = 0 \)
\( (k - 3)(k - 4) = 0 \)
Thus k = 3 or k = 4.

When k = 3, the equation becomes x² - 6x + 9 = 0:
\( x = \frac{6 \pm \sqrt{36 - 36}}{2} = 3 \)
Both roots equal 3.

When k = 4, the equation becomes x² - 8x + 16 = 0:
\( x = \frac{8 \pm \sqrt{64 - 64}}{2} = 4 \)
Both roots equal 4.
In simple words: When the discriminant is zero, you get matching roots. For this equation, k can be 3 or 4. When k = 3, the repeated root is 3; when k = 4, the repeated root is 4.

Exam Tip: Factor the resulting k equation carefully. Both values of k are valid - state both with their corresponding roots clearly.

Question 8. Find the value(s) of p for which the quadratic equation (2p + 1)x² - (7p + 2)x + (7p - 3) = 0 has equal roots. Also find these roots.
Answer: For the equation (2p + 1)x² - (7p + 2)x + (7p - 3) = 0, we identify a = 2p + 1, b = -(7p + 2), c = 7p - 3. Setting discriminant = 0:
\( [-(7p + 2)]^2 - 4(2p + 1)(7p - 3) = 0 \)
\( (7p + 2)^2 - 4(2p + 1)(7p - 3) = 0 \)
\( 49p^2 + 28p + 4 - 4(14p^2 - 6p + 7p - 3) = 0 \)
\( 49p^2 + 28p + 4 - 56p^2 + 24p - 28p + 12 = 0 \)
\( -7p^2 + 24p + 16 = 0 \)
\( 7p^2 - 24p - 16 = 0 \)
\( (7p + 4)(p - 4) = 0 \)
Thus \( p = -\frac{4}{7} \) or p = 4.

When \( p = -\frac{4}{7} \), the equation becomes \( -\frac{1}{7}x^2 + 2x - 7 = 0 \), which simplifies to x² - 14x + 49 = 0:
\( (x - 7)^2 = 0 \)
Both roots equal 7.

When p = 4, the equation becomes 9x² - 30x + 25 = 0:
\( (3x - 5)^2 = 0 \)
Both roots equal \( \frac{5}{3} \).
In simple words: For equal roots, the discriminant must be zero. This gives two values of p. Each p value produces an equation with a perfect square trinomial, giving the repeated root shown above.

Exam Tip: When expanding and simplifying the discriminant condition, factor carefully. Always verify that the final equation is indeed a perfect square.

Question 9. Find the value(s) of p for which the quadratic equation 2x² + 3x + p = 0 has real roots.
Answer: For real roots, we need discriminant \( \geq \) 0. For the equation 2x² + 3x + p = 0, we have a = 2, b = 3, c = p. Applying the condition:
\( b^2 - 4ac \geq 0 \)
\( 9 - 8p \geq 0 \)
\( 9 \geq 8p \)
\( p \leq \frac{9}{8} \)
The equation has real roots whenever \( p \leq \frac{9}{8} \).
In simple words: For real roots, the discriminant must not be negative. This gives us the inequality \( p \leq \frac{9}{8} \), meaning p can be any value up to and including \( \frac{9}{8} \).

Exam Tip: Remember the discriminant inequality for real roots: \( b^2 - 4ac \geq 0 \). Always express your final answer as a range or inequality.

Question 10. Find the least positive value of k for which the equation x² + kx + 4 = 0 has real roots.
Answer: For real roots, the discriminant must satisfy \( b^2 - 4ac \geq 0 \). With a = 1, b = k, c = 4:
\( k^2 - 16 \geq 0 \)
\( k^2 \geq 16 \)
\( (k - 4)(k + 4) \geq 0 \)
This inequality holds when \( k \leq -4 \) or \( k \geq 4 \). Since we need the least positive value, the answer is k = 4.
In simple words: The discriminant being non-negative gives us two ranges: k at most -4, or k at least 4. Among positive values, the smallest is 4.

Exam Tip: Solving the quadratic inequality carefully determines the ranges. Always identify which range contains positive values when the question asks for the least positive value.

Question 11. Find the values of p for which the equation 3x² - px + 5 = 0 has real roots.
Answer: For real roots, we require \( b^2 - 4ac \geq 0 \). With a = 3, b = -p, c = 5:
\( (-p)^2 - 4(3)(5) \geq 0 \)
\( p^2 - 60 \geq 0 \)
\( (p + \sqrt{60})(p - \sqrt{60}) \geq 0 \)
\( (p + 2\sqrt{15})(p - 2\sqrt{15}) \geq 0 \)
This holds when \( p \leq -2\sqrt{15} \) or \( p \geq 2\sqrt{15} \).
In simple words: For the equation to have real roots, p must either be at most \( -2\sqrt{15} \) or at least \( 2\sqrt{15} \). These are the boundaries where the discriminant equals zero.

Exam Tip: When simplifying square roots, factor out perfect squares: \( \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15} \). Express the final answer using simplified radicals.

Exercise 5.5

Question 1(i). Find two consecutive natural numbers such that the sum of their squares is 61.
Answer: Let the two numbers be x and x + 1. According to the given condition:
\( x^2 + (x + 1)^2 = 61 \)
\( x^2 + x^2 + 2x + 1 = 61 \)
\( 2x^2 + 2x + 1 = 61 \)
\( 2x^2 + 2x - 60 = 0 \)
\( x^2 + x - 30 = 0 \)
\( (x + 6)(x - 5) = 0 \)
This gives x = -6 or x = 5. Since we need natural numbers, x = 5. Therefore the two numbers are 5 and 6.
In simple words: Set up an equation with the two numbers as x and x + 1. After simplifying and factoring, only the positive solution gives natural numbers, which are 5 and 6.

Exam Tip: Always check that your solution satisfies the constraint (natural numbers here means positive). Reject negative solutions when the problem specifies a particular type of number.

Question 1(ii). Find two consecutive integers such that the sum of their squares is 61.
Answer: Let the two consecutive integers be x and x + 1. Setting up the equation:
\( x^2 + (x + 1)^2 = 61 \)
\( x^2 + x^2 + 2x + 1 = 61 \)
\( 2x^2 + 2x - 60 = 0 \)
\( x^2 + x - 30 = 0 \)
\( (x + 6)(x - 5) = 0 \)
This yields x = -6 or x = 5. Both are valid integers. When x = 5, the integers are 5 and 6. When x = -6, the integers are -6 and -5. Both pairs satisfy the condition.
In simple words: The equation produces two solutions. Unlike the natural number case, both negative and positive consecutive integers are allowed, so both solutions are valid: either 5 and 6, or -6 and -5.

Exam Tip: When no restriction is placed on whether numbers must be positive, accept all solutions from the quadratic. State both pairs clearly in your answer.

Question 2(i). If the product of two positive consecutive even integers is 288, find the integers.
Answer: Let the two consecutive even integers be x and x + 2. From the given condition:
\( x(x + 2) = 288 \)
\( x^2 + 2x = 288 \)
\( x^2 + 2x - 288 = 0 \)
\( (x - 16)(x + 18) = 0 \)
This gives x = 16 or x = -18. Since we need positive integers, x = 16. Therefore the two integers are 16 and 18.
In simple words: Write the product equation for consecutive even numbers. Factoring gives two solutions, but only the positive one is valid here. The answer is 16 and 18.

Exam Tip: Consecutive even numbers differ by 2, not 1. Verify your final answer by multiplying: 16 × 18 = 288.

Question 2(ii). If the product of two consecutive even integers is 224, find the integers.
Answer: Let the two consecutive even integers be x and x + 2. From the condition:
\( x(x + 2) = 224 \)
\( x^2 + 2x = 224 \)
\( x^2 + 2x - 224 = 0 \)
\( (x - 14)(x + 16) = 0 \)
This yields x = 14 or x = -16. Both are valid since the problem allows any integers. When x = 14, the integers are 14 and 16. When x = -16, the integers are -16 and -14.
In simple words: The equation factors to give x = 14 or x = -16. Since there is no restriction to positive integers, both solutions are acceptable: either 14 and 16, or -16 and -14.

Exam Tip: Always state both pairs when no sign restriction is given. Check both: 14 × 16 = 224 and (-16) × (-14) = 224.

Question 2(iii). Find two consecutive even natural numbers such that the sum of their squares is 340.
Answer: Let the two consecutive even numbers be x and x + 2. According to the condition:
\( x^2 + (x + 2)^2 = 340 \)
\( x^2 + x^2 + 4x + 4 = 340 \)
\( 2x^2 + 4x + 4 = 340 \)
\( 2x^2 + 4x - 336 = 0 \)
\( x^2 + 2x - 168 = 0 \)
\( (x - 12)(x + 14) = 0 \)
This gives x = 12 or x = -14. Since we need natural numbers, x = 12. Therefore the two consecutive even natural numbers are 12 and 14.
In simple words: Set up the sum of squares equation for consecutive even numbers. The factored form yields two solutions, but only the positive one meets the natural number requirement. The answer is 12 and 14.

Exam Tip: Verify: 12² + 14² = 144 + 196 = 340. Always check that your answer satisfies the original condition.

Question 2(iv). Find two consecutive odd integers such that sum of their squares is 394.
Answer: Suppose the two consecutive odd integers are x and x + 2. We are told that the sum of their squares equals 394.

\( x^2 + (x + 2)^2 = 394 \)

\( \Rightarrow x^2 + x^2 + 4 + 4x = 394 \)

\( \Rightarrow 2x^2 + 4x + 4 - 394 = 0 \)

\( \Rightarrow 2x^2 + 4x - 390 = 0 \)

\( \Rightarrow 2(x^2 + 2x - 195) = 0 \)

\( \Rightarrow x^2 + 2x - 195 = 0 \)

\( \Rightarrow x^2 + 15x - 13x - 195 = 0 \)

\( \Rightarrow x(x + 15) - 13(x + 15) = 0 \)

\( \Rightarrow (x + 15)(x - 13) = 0 \) (Factorising left side)

\( \Rightarrow x + 15 = 0 \text{ or } x - 13 = 0 \) (Zero-product rule)

\( \Rightarrow x = -15 \text{ or } x = 13 \)

When x = -15, x + 2 = -13; when x = 13, x + 2 = 15.

Therefore, the required integers are -15, -13 or 13, 15.
In simple words: We find two odd numbers that come one after the other. When we square each and add them together, we get 394. Setting up the equation and solving gives us two pairs of answers.

Exam Tip: Always check both solutions by substituting back into the original condition. Remember that consecutive odd numbers differ by 2, not 1.

Question 3(i). The sum of two numbers is 9 and the sum of their squares is 41. Find the numbers.
Answer: Let one number be x. Since their sum is 9, the other number is 9 - x. We know that the sum of their squares equals 41.

\( \Rightarrow x^2 + (9 - x)^2 = 41 \)

\( \Rightarrow x^2 + x^2 + 81 - 18x = 41 \)

\( \Rightarrow 2x^2 - 18x + 81 - 41 = 0 \)

\( \Rightarrow 2x^2 - 18x + 40 = 0 \)

\( \Rightarrow 2(x^2 - 9x + 20) = 0 \)

\( \Rightarrow x^2 - 9x + 20 = 0 \)

\( \Rightarrow x^2 - 4x - 5x + 20 = 0 \)

\( \Rightarrow x(x - 4) - 5(x - 4) = 0 \)

\( \Rightarrow (x - 4)(x - 5) = 0 \)

\( \Rightarrow x - 4 = 0 \text{ or } x - 5 = 0 \)

\( \Rightarrow x = 4 \text{ or } x = 5 \)

When x = 5, the other number is 9 - 5 = 4. When x = 4, the other number is 9 - 4 = 5.

Therefore, the numbers are 4 and 5.
In simple words: Two numbers add up to 9. Their squares also add up to 41. We solve by writing one number as x and the other as 9 - x, then solve the equation.

Exam Tip: Verify your answer by checking: 4 + 5 = 9 and 4² + 5² = 16 + 25 = 41. This double-check confirms both conditions are met.

Question 3(ii). The difference of two natural numbers is 7 and their product is 450. Find the numbers.
Answer: Let the first number be x. Since their difference is 7, the other number is x + 7. We are given that their product is 450.

\( \Rightarrow x(x + 7) = 450 \)

\( \Rightarrow x^2 + 7x = 450 \)

\( \Rightarrow x^2 + 7x - 450 = 0 \)

\( \Rightarrow x^2 + 25x - 18x - 450 = 0 \)

\( \Rightarrow x(x + 25) - 18(x + 25) = 0 \)

\( \Rightarrow (x + 25)(x - 18) = 0 \)

\( \Rightarrow x + 25 = 0 \text{ or } x - 18 = 0 \)

\( \Rightarrow x = -25 \text{ or } x = 18 \)

Since the numbers are natural numbers, x cannot be -25. Therefore x = 18, and x + 7 = 25.

Hence, the numbers are 18 and 25.
In simple words: Two numbers differ by 7 and multiply to give 450. Setting up and solving the equation gives 18 and 25, both positive whole numbers.

Exam Tip: Always reject negative solutions when the question specifies natural numbers. Verify: 25 - 18 = 7 and 18 × 25 = 450.

Question 4. Five times a certain whole number is equal to three less than twice the square of the number. Find the number.
Answer: Let the number be x. We are told that 5 times the number equals 3 less than twice its square.

\( \Rightarrow 5x = 2x^2 - 3 \)

\( \Rightarrow 5x - 2x^2 + 3 = 0 \)

\( \Rightarrow 2x^2 - 5x - 3 = 0 \) (multiplying by -1)

\( \Rightarrow 2x^2 - 6x + x - 3 = 0 \)

\( \Rightarrow 2x(x - 3) + 1(x - 3) = 0 \)

\( \Rightarrow (2x + 1)(x - 3) = 0 \) (Factorising left side)

\( \Rightarrow 2x + 1 = 0 \text{ or } x - 3 = 0 \) (Zero-product rule)

\( \Rightarrow x = -\frac{1}{2} \text{ or } x = 3 \)

Since the number is a whole number, x cannot be -1/2. Therefore x = 3.

Hence, the required whole number is 3.
In simple words: We need a number where 5 times it equals twice its square minus 3. Solving the resulting quadratic gives 3 as our answer.

Exam Tip: Check that the solution fits the original statement: 5(3) = 2(3²) - 3 gives 15 = 18 - 3 = 15. Also, always reject non-whole number solutions when the question specifies whole numbers.

Question 5. Sum of two natural numbers is 8 and the difference of their reciprocal is \( \frac{2}{15} \). Find the numbers.
Answer: Let the first number be x. Since their sum is 8, the other number is 8 - x. We know that the difference of their reciprocals is 2/15.

\( \Rightarrow \frac{1}{x} - \frac{1}{8-x} = \frac{2}{15} \)

\( \Rightarrow \frac{8-x-x}{x(8-x)} = \frac{2}{15} \) (On taking L.C.M.)

\( \Rightarrow 15(8 - 2x) = 2x(8 - x) \) (On cross multiplication)

\( \Rightarrow 120 - 30x = 16x - 2x^2 \)

\( \Rightarrow 120 - 30x - 16x + 2x^2 = 0 \)

\( \Rightarrow 2x^2 - 46x + 120 = 0 \)

\( \Rightarrow 2(x^2 - 23x + 60) = 0 \)

\( \Rightarrow x^2 - 23x + 60 = 0 \)

\( \Rightarrow x^2 - 20x - 3x + 60 = 0 \)

\( \Rightarrow x(x - 20) - 3(x - 20) = 0 \)

\( \Rightarrow (x - 3)(x - 20) = 0 \) (Factorising left side)

\( \Rightarrow x - 3 = 0 \text{ or } x - 20 = 0 \) (Zero-product rule)

\( \Rightarrow x = 3 \text{ or } x = 20 \)

When x = 20, the other number is 8 - 20 = -12. Since both must be natural numbers, x ≠ 20. Therefore x = 3 and 8 - x = 5.

Hence, the required natural numbers are 3 and 5.
In simple words: Two positive whole numbers add to 8. The difference between 1 divided by the first and 1 divided by the second is 2/15. Solving gives us 3 and 5.

Exam Tip: When working with reciprocals, always check that both solutions make sense in context. Here, one solution gives a negative number, which must be rejected for natural numbers.

Question 6. The difference of the squares of two numbers is 45. The square of the smaller number is 4 times the larger number. Determine the numbers.
Answer: Let the larger number be x. We are told that the square of the smaller number is 4 times the larger number, so the square of the smaller number = 4x. This means the smaller number = \( \sqrt{4x} \).

The difference of their squares is 45:

\( \Rightarrow x^2 - 4x = 45 \)

\( \Rightarrow x^2 - 4x - 45 = 0 \)

\( \Rightarrow x^2 - 9x + 5x - 45 = 0 \)

\( \Rightarrow x(x - 9) + 5(x - 9) = 0 \)

\( \Rightarrow (x + 5)(x - 9) = 0 \)

\( \Rightarrow x + 5 = 0 \text{ or } x - 9 = 0 \)

\( \Rightarrow x = -5 \text{ or } x = 9 \)

When x = -5, \( \sqrt{4x} = \sqrt{-20} \), which has no real value. Therefore x = 9, and \( \sqrt{4x} = \sqrt{36} = 6 \) or -6.

Hence, the two numbers are 9, 6 or 9, -6.
In simple words: We have two numbers where one squared minus the other squared gives 45, and the smaller squared equals 4 times the larger. Solving the equation gives 9 as the larger number and 6 or -6 as the smaller.

Exam Tip: When a square root appears in your working, check whether negative solutions are valid. Here, x = -5 leads to an impossible square root, so it must be rejected.

Question 7. There are three consecutive positive integers such that the sum of the square of the first and the product of the other two is 154. What are the integers?
Answer: Let the three consecutive integers be x, x + 1, and x + 2. We are told that the sum of the square of the first and the product of the other two is 154.

\( \Rightarrow x^2 + (x + 1)(x + 2) = 154 \)

\( \Rightarrow x^2 + x^2 + 2x + x + 2 = 154 \)

\( \Rightarrow 2x^2 + 3x + 2 - 154 = 0 \)

\( \Rightarrow 2x^2 + 3x - 152 = 0 \)

\( \Rightarrow 2x^2 + 19x - 16x - 152 = 0 \)

\( \Rightarrow x(2x + 19) - 8(2x + 19) = 0 \)

\( \Rightarrow (x - 8)(2x + 19) = 0 \)

\( \Rightarrow x - 8 = 0 \text{ or } 2x + 19 = 0 \)

\( \Rightarrow x = 8 \text{ or } x = -\frac{19}{2} \)

Since the integers are positive, x = 8. Therefore x + 1 = 9 and x + 2 = 10.

Hence, the required numbers are 8, 9, 10.
In simple words: Three numbers come one after another. The first squared plus the product of the other two equals 154. Setting up and solving gives 8, 9, and 10.

Exam Tip: Always reject fractional or negative solutions when the question asks for positive integers. Verify your answer: 8² + (9 × 10) = 64 + 90 = 154.

Question 8(i). Find three successive even natural numbers, the sum of whose squares is 308.
Answer: Let the three required numbers be x, x + 2, and x + 4. We are given that the sum of their squares equals 308.

\( \Rightarrow x^2 + (x + 2)^2 + (x + 4)^2 = 308 \)

\( \Rightarrow x^2 + x^2 + 4 + 4x + x^2 + 16 + 8x = 308 \)

\( \Rightarrow 3x^2 + 12x + 20 = 308 \)

\( \Rightarrow 3x^2 + 12x - 288 = 0 \)

\( \Rightarrow 3(x^2 + 4x - 96) = 0 \)

\( \Rightarrow x^2 + 4x - 96 = 0 \)

\( \Rightarrow x^2 + 12x - 8x - 96 = 0 \)

\( \Rightarrow x(x + 12) - 8(x + 12) = 0 \)

\( \Rightarrow (x - 8)(x + 12) = 0 \)

\( \Rightarrow x - 8 = 0 \text{ or } x + 12 = 0 \)

\( \Rightarrow x = 8 \text{ or } x = -12 \)

Since the numbers are natural, x = 8. Therefore x + 2 = 10 and x + 4 = 12.

Hence, the required numbers are 8, 10, 12.
In simple words: Three even numbers in a row have squares that sum to 308. Solving the equation gives 8, 10, and 12.

Exam Tip: For consecutive even numbers, remember they differ by 2, not 1. Verify: 8² + 10² + 12² = 64 + 100 + 144 = 308.

Question 8(ii). Find three consecutive odd integers, the sum of whose squares is 83.
Answer: Let the three required numbers be x, x + 2, and x + 4. We are told that the sum of their squares equals 83.

\( \Rightarrow x^2 + (x + 2)^2 + (x + 4)^2 = 83 \)

\( \Rightarrow x^2 + x^2 + 4 + 4x + x^2 + 16 + 8x = 83 \)

\( \Rightarrow 3x^2 + 12x + 20 = 83 \)

\( \Rightarrow 3x^2 + 12x - 63 = 0 \)

\( \Rightarrow 3(x^2 + 4x - 21) = 0 \)

\( \Rightarrow x^2 + 4x - 21 = 0 \)

\( \Rightarrow x^2 + 7x - 3x - 21 = 0 \)

\( \Rightarrow x(x + 7) - 3(x + 7) = 0 \)

\( \Rightarrow (x + 7)(x - 3) = 0 \)

\( \Rightarrow x + 7 = 0 \text{ or } x - 3 = 0 \)

\( \Rightarrow x = -7 \text{ or } x = 3 \)

When x = -7, the numbers are -7, -5, -3. When x = 3, the numbers are 3, 5, 7.

Hence, the required numbers are -7, -5, -3 and 3, 5, 7.
In simple words: Three odd numbers in a row have squares that sum to 83. We get two sets of answers: negative odd numbers or positive odd numbers.

Exam Tip: Both solutions are valid here since the question does not specify positive integers. Always check: (-7)² + (-5)² + (-3)² = 49 + 25 + 9 = 83 and 3² + 5² + 7² = 9 + 25 + 49 = 83.

Question 9. In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from both the numerator and denominator, the fraction is decreased by \( \frac{1}{14} \). Find the fraction.
Answer: Let the numerator be x. Since the denominator is greater than the numerator by 3, the denominator is x + 3. The fraction is \( \frac{x}{x+3} \). When 1 is subtracted from both, the new fraction is \( \frac{x-1}{x+2} \). We are told this new fraction is \( \frac{1}{14} \) less than the original:

\( \Rightarrow \frac{x}{x+3} - \frac{x-1}{x+2} = \frac{1}{14} \)

\( \Rightarrow \frac{x(x+2) - (x-1)(x+3)}{(x+3)(x+2)} = \frac{1}{14} \) (On taking L.C.M.)

\( \Rightarrow \frac{x^2 + 2x - (x^2 + 3x - x - 3)}{x^2 + 2x + 3x + 6} = \frac{1}{14} \)

\( \Rightarrow \frac{x^2 + 2x - x^2 - 3x + x + 3}{x^2 + 5x + 6} = \frac{1}{14} \)

\( \Rightarrow \frac{3}{x^2 + 5x + 6} = \frac{1}{14} \)

\( \Rightarrow 3 \times 14 = x^2 + 5x + 6 \) (Cross multiplying)

\( \Rightarrow x^2 + 5x + 6 = 42 \)

\( \Rightarrow x^2 + 5x + 6 - 42 = 0 \)

\( \Rightarrow x^2 + 5x - 36 = 0 \)

\( \Rightarrow x^2 + 9x - 4x - 36 = 0 \)

\( \Rightarrow x(x + 9) - 4(x + 9) = 0 \)

\( \Rightarrow (x - 4)(x + 9) = 0 \)

\( \Rightarrow x = 4 \text{ or } x = -9 \)

When x = -9, the fraction is \( \frac{-9}{-6} = \frac{3}{2} \), which has the numerator greater than the denominator. This contradicts the given condition. Therefore x = 4, and the fraction is \( \frac{4}{7} \).

Hence, the fraction is \( \frac{4}{7} \).
In simple words: The bottom number is 3 more than the top number. When we take 1 away from both, the fraction gets smaller by 1/14. Solving this equation gives 4/7.

Exam Tip: Always verify that your solution satisfies all given conditions. Here, check: 4/7 - 3/6 = 4/7 - 1/2 = 8/14 - 7/14 = 1/14. Also reject solutions that contradict the problem statement.

Question 10. The sum of numerator and denominator of a certain positive fraction is 8. If 2 is added to both the numerator and denominator, the fraction is increased by \( \frac{4}{35} \). Find the fraction.
Answer: Let the denominator be x, so the numerator is 8 - x. The fraction is \( \frac{8-x}{x} \). When 2 is added to both, the new fraction is \( \frac{8-x+2}{x+2} = \frac{10-x}{x+2} \). We are told this new fraction is \( \frac{4}{35} \) more than the original:

\( \Rightarrow \frac{10-x}{x+2} - \frac{8-x}{x} = \frac{4}{35} \)

\( \Rightarrow \frac{x(10-x) - (x+2)(8-x)}{x(x+2)} = \frac{4}{35} \) (On taking L.C.M.)

\( \Rightarrow \frac{10x - x^2 - (8x - x^2 + 16 - 2x)}{x^2 + 2x} = \frac{4}{35} \)

\( \Rightarrow \frac{10x - x^2 - 8x + x^2 - 16 + 2x}{x^2 + 2x} = \frac{4}{35} \)

\( \Rightarrow \frac{4x - 16}{x^2 + 2x} = \frac{4}{35} \)

\( \Rightarrow 35(4x - 16) = 4(x^2 + 2x) \)

\( \Rightarrow 140x - 560 = 4x^2 + 8x \)

\( \Rightarrow 4x^2 + 8x - 140x + 560 = 0 \)

\( \Rightarrow 4x^2 - 132x + 560 = 0 \)

\( \Rightarrow 4(x^2 - 33x + 140) = 0 \)

\( \Rightarrow x^2 - 33x + 140 = 0 \)

\( \Rightarrow x^2 - 28x - 5x + 140 = 0 \)

\( \Rightarrow x(x - 28) - 5(x - 28) = 0 \)

\( \Rightarrow (x - 28)(x - 5) = 0 \)

\( \Rightarrow x = 28 \text{ or } x = 5 \)

When x = 28, the numerator is 8 - 28 = -20, making the fraction negative. This is not a positive fraction, so x ≠ 28. Therefore x = 5, and the numerator is 8 - 5 = 3. The fraction is \( \frac{3}{5} \).

Hence, the fraction is \( \frac{3}{5} \).
In simple words: The top and bottom numbers add to 8. When we add 2 to both, the fraction gets larger by 4/35. Solving gives 3/5.

Exam Tip: Always ensure that your solution is valid based on the problem constraints. Here, one solution produces a negative fraction, which must be rejected since we need a positive fraction.

Question 11. A two digit number contains the Larger digit at ten's place. The product of the digits is 27 and the difference between two digits is 6. Find the number.
Answer: Let the unit's digit be x. Since the larger digit is at the ten's place and the difference between the digits is 6, the ten's digit is x + 6. The number formed is \( 10(x + 6) + x = 10x + 60 + x = 11x + 60 \). We are told that the product of the digits is 27:

\( \Rightarrow x(x + 6) = 27 \)

\( \Rightarrow x^2 + 6x = 27 \)

\( \Rightarrow x^2 + 6x - 27 = 0 \)

\( \Rightarrow x^2 + 9x - 3x - 27 = 0 \)

\( \Rightarrow x(x + 9) - 3(x + 9) = 0 \)

\( \Rightarrow (x - 3)(x + 9) = 0 \)

\( \Rightarrow x = 3 \text{ or } x = -9 \)

Since x is a digit, it must be non-negative, so x = 3. The ten's digit is 3 + 6 = 9. The number is 93.

Hence, the required number is 93.
In simple words: A two-digit number has its larger digit in the tens place. The two digits multiply to 27 and differ by 6. This gives us 9 and 3, forming the number 93.

Exam Tip: Remember that digits must be between 0 and 9, so reject any solution giving a digit outside this range. Verify: 9 × 3 = 27 and 9 - 3 = 6.

Question 12. A two digit positive number is such that the product of its digit is 6. If 9 is added to the number, the digits interchange their place. Find the number.
Answer: Let the digit at the unit's place be x. Since the product of digits is 6, the digit in the ten's place is \( \frac{6}{x} \).

The number can be written as \( 10 \times \frac{6}{x} + x = \frac{60}{x} + x = \frac{60 + x^2}{x} \).

When the digits swap positions, the new number becomes \( 10 \times x + \frac{6}{x} = 10x + \frac{6}{x} \).

According to the given condition:
\( \frac{x^2 + 60}{x} + 9 = 10x + \frac{6}{x} \)

Simplifying:
\( \frac{x^2 + 60 + 9x}{x} = \frac{10x^2 + 6}{x} \)

Cross-multiplying:
\( x^2 + 9x + 60 = 10x^2 + 6 \)

\( 9x^2 - 9x - 54 = 0 \)

\( x^2 - x - 6 = 0 \)

\( (x + 2)(x - 3) = 0 \)

Therefore, \( x = 3 \) or \( x = -2 \). Since the number must be positive, \( x = 3 \).

The number is \( \frac{3^2 + 60}{3} = \frac{69}{3} = 23 \).
In simple words: The digit at the unit's place is 3 and the digit at the ten's place is 2, making the number 23. When you add 9 to this number, you get 32, which is 23 with its digits reversed.

Exam Tip: Always check that your solution satisfies all the given conditions - verify that the product of digits equals 6 and that adding 9 to the original number gives the digit-reversed form.

Question 13. A rectangle of area 105 cm² has its length equal to x cm. Write down its breadth in terms of x. Given that the perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of rectangle.
Answer: Length of the rectangle = x cm.

Since the area is 105 cm², the breadth = \( \frac{105}{x} \) cm.

The perimeter of a rectangle is given by:
\( \text{Perimeter} = 2(\text{length} + \text{breadth}) \)

\( 2\left(x + \frac{105}{x}\right) = 44 \)

Dividing both sides by 2:
\( x + \frac{105}{x} = 22 \)

Multiplying by x:
\( x^2 + 105 = 22x \)

\( x^2 - 22x + 105 = 0 \)

\( (x - 7)(x - 15) = 0 \)

Therefore, \( x = 7 \) or \( x = 15 \).

When \( x = 7 \): breadth = \( \frac{105}{7} = 15 \) cm, so length = 7 cm and breadth = 15 cm.

When \( x = 15 \): breadth = \( \frac{105}{15} = 7 \) cm, so length = 15 cm and breadth = 7 cm.

The dimensions of the rectangle are length = 15 cm and breadth = 7 cm.
In simple words: The length and breadth are 15 cm and 7 cm. It doesn't matter which one you call the length and which the breadth - the rectangle is the same either way.

Exam Tip: When you get two values for x, remember that they simply represent different orientations of the same rectangle. State both solutions and indicate the final dimensions clearly.

Question 14. A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.
Answer: The garden has dimensions 10 m by 16 m. When surrounded by a concrete walk of width x, the outer dimensions become (10 + 2x) m by (16 + 2x) m (x is added on both sides).

Area of garden and walk combined = \( (10 + 2x)(16 + 2x) \) m²

Area of walk = Area of combined - Area of garden

\( (10 + 2x)(16 + 2x) - 10 \times 16 = 120 \)

Expanding:
\( 160 + 20x + 32x + 4x^2 - 160 = 120 \)

\( 4x^2 + 52x = 120 \)

\( 4x^2 + 52x - 120 = 0 \)

\( x^2 + 13x - 30 = 0 \)

\( (x + 15)(x - 2) = 0 \)

Therefore, \( x = 2 \) or \( x = -15 \). Since width cannot be negative, \( x = 2 \) m.
In simple words: The concrete walk around the garden must be 2 metres wide. This adds a strip of concrete all around the garden on all sides.

Exam Tip: Remember that when a uniform walk surrounds a rectangle, the width is added on both opposite sides, so the outer dimensions increase by 2x in each direction (x on each side).

Question 15. The length of a rectangle exceeds its breadth by 5m. If the breadth were doubled and the length reduced by 9m, the area of the rectangle would have increased by 140 m². Find its dimensions.
Answer: Let the breadth of the rectangle be x metres. Then the length is (x + 5) metres.

Original area = \( x(x + 5) = x^2 + 5x \) m².

If breadth is doubled: new breadth = 2x metres. If length is reduced by 9m: new length = (x + 5 - 9) = (x - 4) metres.

New area = \( (x - 4)(2x) = 2x^2 - 8x \) m².

According to the given condition, the area increased by 140 m²:
\( 2x^2 - 8x - (x^2 + 5x) = 140 \)

\( 2x^2 - 8x - x^2 - 5x = 140 \)

\( x^2 - 13x = 140 \)

\( x^2 - 13x - 140 = 0 \)

\( (x + 7)(x - 20) = 0 \)

Therefore, \( x = -7 \) or \( x = 20 \). Since breadth cannot be negative, \( x = 20 \) metres.

Length = \( x + 5 = 20 + 5 = 25 \) metres.
In simple words: The breadth is 20 metres and the length is 25 metres. When you double the breadth to 40 metres and reduce the length to 16 metres, the area becomes much larger by exactly 140 square metres.

Exam Tip: Carefully track what happens to the area when dimensions change. Set up the equation by comparing the new area to the old area, not just to an absolute value.

Question 16. The perimeter of a rectangular plot is 180 m and its area is 1800 m². Take the length of the plot as x metres. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the values of length, breadth and the area to write an equation in x. Solve the equation to calculate the length and breadth of the plot.
Answer: Let length = x metres.

From the perimeter:
\( 2(x + \text{breadth}) = 180 \)

\( x + \text{breadth} = 90 \)

\( \text{breadth} = 90 - x \) metres.

From the area condition:
\( x(90 - x) = 1800 \)

\( 90x - x^2 = 1800 \)

\( x^2 - 90x + 1800 = 0 \)

\( (x - 30)(x - 60) = 0 \)

Therefore, \( x = 30 \) or \( x = 60 \).

When \( x = 30 \): breadth = \( 90 - 30 = 60 \) m, so length = 30 m and breadth = 60 m.

When \( x = 60 \): breadth = \( 90 - 60 = 30 \) m, so length = 60 m and breadth = 30 m.

The dimensions are length = 60 m and breadth = 30 m.
In simple words: The plot is either 60 metres long and 30 metres wide, or 30 metres long and 60 metres wide - it's the same rectangle, just named differently.

Exam Tip: When solving rectangular problems, use the perimeter to express one dimension in terms of the other, then substitute into the area equation to get a single variable equation.

Question 17. The lengths of parallel sides of a trapezium are (x + 9) cm and (2x - 3) cm, and the distance between them is (x + 4) cm. If its area is 540 cm², find x.
Answer: The area of a trapezium is given by:
\( \text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{distance between them}) \)

\( \frac{1}{2} \times [(x + 9) + (2x - 3)] \times (x + 4) = 540 \)

\( \frac{1}{2} \times (3x + 6) \times (x + 4) = 540 \)

\( (3x + 6)(x + 4) = 1080 \)

\( 3x^2 + 12x + 6x + 24 = 1080 \)

\( 3x^2 + 18x + 24 = 1080 \)

\( 3x^2 + 18x - 1056 = 0 \)

\( x^2 + 6x - 352 = 0 \)

\( (x + 22)(x - 16) = 0 \)

Therefore, \( x = -22 \) or \( x = 16 \).

Since the side lengths must be positive, we check: when \( x = 16 \), the sides are \( 16 + 9 = 25 \) cm and \( 2(16) - 3 = 29 \) cm, both positive. When \( x = -22 \), the sides become negative, so this is rejected.

Thus, \( x = 16 \).
In simple words: The value of x is 16. This makes the two parallel sides 25 cm and 29 cm long, with a perpendicular distance of 20 cm between them.

Exam Tip: Always check that your solution makes geometric sense - all side lengths and distances must be positive. Reject any solution that produces negative dimensions.

Question 18. If the perimeter of a rectangular plot is 68 m and length of its diagonal is 26 m, find its area.
Answer: Let length = l and breadth = b.

From the perimeter:
\( 2(l + b) = 68 \)

\( l + b = 34 \)

\( l = 34 - b \) ... (Equation A)

For a rectangle, the diagonal d satisfies:
\( d^2 = l^2 + b^2 \)

Given diagonal = 26 m:
\( l^2 + b^2 = 26^2 = 676 \)

Substituting \( l = 34 - b \):
\( (34 - b)^2 + b^2 = 676 \)

\( 1156 - 68b + b^2 + b^2 = 676 \)

\( 2b^2 - 68b + 1156 - 676 = 0 \)

\( 2b^2 - 68b + 480 = 0 \)

\( b^2 - 34b + 240 = 0 \)

\( (b - 24)(b - 10) = 0 \)

Therefore, \( b = 24 \) or \( b = 10 \).

When \( b = 24 \): \( l = 34 - 24 = 10 \) m. When \( b = 10 \): \( l = 34 - 10 = 24 \) m.

In either case, the dimensions are 24 m and 10 m.

Area = \( 24 \times 10 = 240 \) m².
In simple words: The rectangle is 24 metres long and 10 metres wide. Its diagonal stretches 26 metres across from one corner to the opposite corner, and its area is 240 square metres.

Exam Tip: For rectangles, remember the relationship between sides and diagonal using the Pythagorean theorem. Setting up two equations from the perimeter and diagonal conditions allows you to solve for both dimensions.

Question 19. If the sum of two smaller sides of a right-angled triangle is 17 cm and the perimeter is 30 cm, then find the area of the triangle.
Answer: Let one of the two smaller sides be x cm. Then the other smaller side is (17 - x) cm.

The hypotenuse = Perimeter - sum of smaller sides = 30 - 17 = 13 cm.

Using the Pythagorean theorem:
\( x^2 + (17 - x)^2 = 13^2 \)

\( x^2 + 289 - 34x + x^2 = 169 \)

\( 2x^2 - 34x + 289 - 169 = 0 \)

\( 2x^2 - 34x + 120 = 0 \)

\( x^2 - 17x + 60 = 0 \)

\( (x - 5)(x - 12) = 0 \)

Therefore, \( x = 5 \) or \( x = 12 \).

The two smaller sides are 5 cm and 12 cm (in either order).

Area = \( \frac{1}{2} \times 5 \times 12 = 30 \) cm².
In simple words: The triangle has two shorter sides of 5 cm and 12 cm, and a longest side (hypotenuse) of 13 cm. It's a right triangle, so its area is half the product of the two shorter sides, which equals 30 square centimetres.

Exam Tip: In right triangle problems, use the Pythagorean theorem to set up your equation. When you get two solutions for the sides, they just represent the same pair of perpendicular sides in different order.

Question 20. The hypotenuse of a grassy land in the shape of a right triangle is 1 metre more than twice the shortest side. If the third side is 7 metres more than the shortest side, find the sides of the grassy land.
Answer: Let the shortest side be x metres.

Then hypotenuse = (2x + 1) metres and the third side = (x + 7) metres.

Using the Pythagorean theorem:
\( (2x + 1)^2 = x^2 + (x + 7)^2 \)

\( 4x^2 + 4x + 1 = x^2 + x^2 + 14x + 49 \)

\( 4x^2 + 4x + 1 = 2x^2 + 14x + 49 \)

\( 2x^2 - 10x - 48 = 0 \)

\( x^2 - 5x - 24 = 0 \)

\( (x - 8)(x + 3) = 0 \)

Therefore, \( x = 8 \) or \( x = -3 \). Since side length cannot be negative, \( x = 8 \) metres.

Shortest side = 8 m, third side = \( 8 + 7 = 15 \) m, hypotenuse = \( 2(8) + 1 = 17 \) m.
In simple words: The three sides of the triangle are 8 metres, 15 metres, and 17 metres. The 17-metre side is the hypotenuse (longest side) opposite the right angle.

Exam Tip: When setting up the Pythagorean theorem, ensure the hypotenuse is on one side and the two perpendicular sides are on the other. Check that your solution satisfies the Pythagorean relationship.

Question 21. Mohini wishes to fit three rods together in the shape of a right triangle. If the hypotenuse is 2 cm longer than the base and 4 cm longer than the shortest side, find the lengths of the rods.
Answer: Let the length of the hypotenuse be x cm.

Then the base = (x - 2) cm and the shortest side = (x - 4) cm.

Using the Pythagorean theorem:
\( x^2 = (x - 2)^2 + (x - 4)^2 \)

\( x^2 = x^2 - 4x + 4 + x^2 - 8x + 16 \)

\( x^2 = 2x^2 - 12x + 20 \)

\( 0 = x^2 - 12x + 20 \)

\( x^2 - 12x + 20 = 0 \)

Using the quadratic formula or factoring, we find this should simplify. Let me recalculate: \( x^2 - 10x - 20 = 0 \) gives us the correct path. Factoring: \( (x - 10)(x + 2) = 0 \) is not correct. Using the quadratic formula yields approximately x ≈ 10.95 or solving differently:

Actually, \( x^2 - 12x + 20 = 0 \) factors as we solve systematically, yielding \( x = 10 \) cm (taking the valid positive solution).

Hypotenuse = 10 cm, base = 10 - 2 = 8 cm, shortest side = 10 - 4 = 6 cm.
In simple words: The three rods have lengths 6 cm, 8 cm, and 10 cm. When arranged with the 10 cm rod as the hypotenuse, the 6 cm and 8 cm rods form a right angle between them.

Exam Tip: Always verify your solution by checking the Pythagorean theorem: \( 6^2 + 8^2 = 36 + 64 = 100 = 10^2 \). This confirms the triangle is valid.

Question 22. The longer side = (x - 2) cm and shortest side = (x - 4) cm. Apply the Pythagorean theorem to find the dimensions of the right-angled triangle.
Answer: Using the Pythagorean theorem, the hypotenuse squared equals the sum of the squares of the other two sides.
\( x^2 = (x - 2)^2 + (x - 4)^2 \)
\( \Rightarrow x^2 = x^2 + 4 - 4x + x^2 + 16 - 8x \)
\( \Rightarrow x^2 = 2x^2 - 12x + 20 \)
\( \Rightarrow 2x^2 - x^2 - 12x + 20 = 0 \)
\( \Rightarrow x^2 - 12x + 20 = 0 \)
\( \Rightarrow x^2 - 10x - 2x + 20 = 0 \)
\( \Rightarrow x(x - 10) - 2(x - 10) = 0 \)
\( \Rightarrow (x - 2)(x - 10) = 0 \)
\( \Rightarrow x = 2 \text{ or } x = 10 \)
Since \( x = 2 \) makes the shortest side negative \( (x - 4) = -2 \), we reject it. If \( x = 10 \), then \( x - 2 = 8 \) and \( x - 4 = 6 \). The triangle's dimensions are: Hypotenuse = 10 cm, Longer side = 8 cm, Shortest side = 6 cm.
In simple words: The three sides of the right-angled triangle are 6 cm, 8 cm, and 10 cm. You can check this using the rule \( 6^2 + 8^2 = 10^2 \), which is \( 36 + 64 = 100 \).

Exam Tip: Always reject any solution that gives a negative side length or distance - such values are not valid in geometry problems.

Question 23. In a P.T. display, 480 students are arranged in rows and columns. If there are 4 more students in each row than the number of rows, find the number of students in each row.
Answer: Let the number of rows be \( x \). Then the number of students in each row is \( x + 4 \). Since the total number of students is 480:
\( x(x + 4) = 480 \)
\( \Rightarrow x^2 + 4x = 480 \)
\( \Rightarrow x^2 + 4x - 480 = 0 \)
\( \Rightarrow x^2 + 24x - 20x - 480 = 0 \)
\( \Rightarrow x(x + 24) - 20(x + 24) = 0 \)
\( \Rightarrow (x - 20)(x + 24) = 0 \)
\( \Rightarrow x = 20 \text{ or } x = -24 \)
Since the number of rows cannot be negative, \( x \neq -24 \). Therefore \( x = 20 \), which means \( x + 4 = 24 \). The number of students in each row is 24.
In simple words: There are 20 rows with 24 students in each row. When you multiply 20 by 24, you get 480 students total.

Exam Tip: Check your answer by substituting back: if rows = 20 and students per row = 24, then 20 × 24 = 480 ✓

Question 24. In an auditorium, the number of rows was equal to number of seats in each row. If the number of rows is doubled and the number of seats in each row is reduced by 5, then the total number of seats is increased by 375. How many rows were there?
Answer: Let the number of rows = number of seats in each row = \( x \). Then the original total number of seats = \( x \times x = x^2 \). After the change, the new total is \( 2x(x - 5) \). Since the increase in total seats is 375:
\( 2x(x - 5) - x^2 = 375 \)
\( \Rightarrow 2x^2 - 10x - x^2 = 375 \)
\( \Rightarrow x^2 - 10x - 375 = 0 \)
\( \Rightarrow x^2 - 25x + 15x - 375 = 0 \)
\( \Rightarrow x(x - 25) + 15(x - 25) = 0 \)
\( \Rightarrow (x - 25)(x + 15) = 0 \)
\( \Rightarrow x = 25 \text{ or } x = -15 \)
Since the number of rows cannot be negative, \( x \neq -15 \). Therefore, there were 25 rows.
In simple words: Originally there were 25 rows and 25 seats per row (625 seats total). After doubling rows to 50 and reducing seats per row to 20, the new total is 1000 seats, which is 375 more than the original.

Exam Tip: For arrangement problems, remember that "number of rows = number of seats per row" means the original setup is square-shaped - this is an important condition to recognize.

Question 25. At an annual function of a school, each student gives gift to every other student. If the number of gifts is 1980, find the number of students.
Answer: Let the number of students be \( x \). Each student gives a gift to \( (x - 1) \) other students. So the total number of gifts is \( x(x - 1) \):
\( x(x - 1) = 1980 \)
\( \Rightarrow x^2 - x = 1980 \)
\( \Rightarrow x^2 - x - 1980 = 0 \)
\( \Rightarrow x^2 - 45x + 44x - 1980 = 0 \)
\( \Rightarrow x(x - 45) + 44(x - 45) = 0 \)
\( \Rightarrow (x - 45)(x + 44) = 0 \)
\( \Rightarrow x = 45 \text{ or } x = -44 \)
Since the number of students cannot be negative, \( x \neq -44 \). Therefore, there are 45 students.
In simple words: With 45 students, each one gives gifts to 44 others. The total gifts = 45 × 44 = 1980.

Exam Tip: In "gift-giving" problems, each student gives to every OTHER student (not to themselves), so one student gives \( (n - 1) \) gifts, and with \( n \) students total, the count is \( n(n - 1) \).

Question 26. A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hours longer to cover the total distance. Assuming the uniform speed to be 'x' km/h, form an equation and solve it to evaluate x.
Answer: Let the uniform speed of the bus be \( x \) km/h. Due to rain, the speed reduces to \( (x - 10) \) km/h. Since time = distance/speed, and the reduced speed takes 2 hours longer:
\( \frac{240}{x - 10} - \frac{240}{x} = 2 \)
\( \Rightarrow \frac{240x - 240(x - 10)}{x(x - 10)} = 2 \)
\( \Rightarrow 240x - 240x + 2400 = 2x(x - 10) \)
\( \Rightarrow 2400 = 2x^2 - 20x \)
\( \Rightarrow 2x^2 - 20x - 2400 = 0 \)
\( \Rightarrow x^2 - 10x - 1200 = 0 \)
\( \Rightarrow x^2 - 40x + 30x - 1200 = 0 \)
\( \Rightarrow x(x - 40) + 30(x - 40) = 0 \)
\( \Rightarrow (x + 30)(x - 40) = 0 \)
\( \Rightarrow x = -30 \text{ or } x = 40 \)
Since speed cannot be negative, \( x \neq -30 \). Therefore, the uniform speed is 40 km/h. The equation is \( \frac{240}{x - 10} - \frac{240}{x} = 2 \)
In simple words: At normal speed of 40 km/h, the bus takes 6 hours. At reduced speed of 30 km/h, it takes 8 hours - which is 2 hours longer.

Exam Tip: Always check: if speed increases, time decreases, and vice versa. Use the time equation carefully - the slower journey takes MORE time.

Question 27. The speed of an express train is x km/h and the speed of an ordinary train is 12 km/h less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of 240 km, find the speed of the express train.
Answer: Let the speed of the express train be \( x \) km/h. Then the speed of the ordinary train is \( (x - 12) \) km/h. Since the ordinary train takes 1 hour longer:
\( \frac{240}{x - 12} - \frac{240}{x} = 1 \)
\( \Rightarrow \frac{240x - 240(x - 12)}{x(x - 12)} = 1 \)
\( \Rightarrow 240x - 240x + 2880 = x(x - 12) \)
\( \Rightarrow 2880 = x^2 - 12x \)
\( \Rightarrow x^2 - 12x - 2880 = 0 \)
\( \Rightarrow x^2 - 60x + 48x - 2880 = 0 \)
\( \Rightarrow x(x - 60) + 48(x - 60) = 0 \)
\( \Rightarrow (x - 60)(x + 48) = 0 \)
\( \Rightarrow x = 60 \text{ or } x = -48 \)
Since speed cannot be negative, \( x \neq -48 \). Therefore, the speed of the express train is 60 km/h.
In simple words: The express train travels at 60 km/h and covers 240 km in 4 hours. The ordinary train at 48 km/h covers the same distance in 5 hours - exactly 1 hour longer.

Exam Tip: Set up the time difference equation carefully: (slower train time) - (faster train time) = time difference given.

Question 28. A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.
Answer: Let the speed of the car be \( x \) km/h. Note that 1 hour 40 minutes = \( \frac{100}{60} = \frac{5}{3} \) hours. Since the increased speed saves \( \frac{5}{3} \) hours:
\( \frac{400}{x} - \frac{400}{x + 12} = \frac{5}{3} \)
\( \Rightarrow 100\left(\frac{4}{x} - \frac{4}{x + 12}\right) = \frac{100}{60} \)
\( \Rightarrow \frac{4}{x} - \frac{4}{x + 12} = \frac{1}{60} \)
\( \Rightarrow \frac{4(x + 12) - 4x}{x(x + 12)} = \frac{1}{60} \)
\( \Rightarrow 60(4x + 48 - 4x) = x(x + 12) \)
\( \Rightarrow 60 \times 48 = x^2 + 12x \)
\( \Rightarrow x^2 + 12x - 2880 = 0 \)
\( \Rightarrow x^2 + 60x - 48x - 2880 = 0 \)
\( \Rightarrow x(x + 60) - 48(x + 60) = 0 \)
\( \Rightarrow (x - 48)(x + 60) = 0 \)
\( \Rightarrow x = 48 \text{ or } x = -60 \)
Since speed cannot be negative, \( x \neq -60 \). Therefore, the original speed is 48 km/h.
In simple words: At 48 km/h, the car takes \( \frac{400}{48} = 8.33 \) hours. At 60 km/h, it takes \( \frac{400}{60} = 6.67 \) hours. The difference is 1 hour 40 minutes.

Exam Tip: Always convert time units carefully - convert minutes to hours by dividing by 60 before setting up equations.

Question 29. An aeroplane covered a distance of 400 km at an average speed of x km/h. On the return journey, the speed was increased by 40 km/h. Write down an expression for the time taken for: (i) the onward journey (ii) the return journey. If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value.
Answer:
(i) For the onward journey: Distance = 400 km, Speed = \( x \) km/h
\( \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{400}{x} \text{ hours} \)

(ii) For the return journey: Distance = 400 km, Speed = \( (x + 40) \) km/h
\( \text{Time} = \frac{400}{x + 40} \text{ hours} \)

Since the return journey took 30 minutes \( = \frac{1}{2} \) hour less:
\( \frac{400}{x} - \frac{400}{x + 40} = \frac{1}{2} \)
\( \Rightarrow \frac{400(x + 40) - 400x}{x(x + 40)} = \frac{1}{2} \)
\( \Rightarrow \frac{400x - 400x + 16000}{x(x + 40)} = \frac{1}{2} \)
\( \Rightarrow 16000 \times 2 = x(x + 40) \)
\( \Rightarrow 32000 = x^2 + 40x \)
\( \Rightarrow x^2 + 40x - 32000 = 0 \)
\( \Rightarrow x^2 + 200x - 160x - 32000 = 0 \)
\( \Rightarrow x(x + 200) - 160(x + 200) = 0 \)
\( \Rightarrow (x - 160)(x + 200) = 0 \)
\( \Rightarrow x = 160 \text{ or } x = -200 \)
Since speed cannot be negative, \( x \neq -200 \). Therefore, \( x = 160 \) km/h.
Equation in x: \( \frac{400}{x} - \frac{400}{x + 40} = \frac{1}{2} \)
In simple words: At 160 km/h, the onward flight takes 2.5 hours. At 200 km/h, the return flight takes 2 hours - which is 30 minutes (half an hour) less.

Exam Tip: For multi-part questions, show both expressions clearly before forming the final equation - examiners give marks for correct expressions even if the final answer is wrong.

Question 30. The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A car travels at a speed of x km/h, and the train travels at a speed which is 16 km/h faster than the car. Calculate: (i) The time taken by car to reach town B from A, in terms of x. (ii) The time taken by the train to reach town B from A, in terms of x. (iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x, and solve it. (iv) Hence, find the speed of the train.
Answer:
(i) Speed of car = \( x \) km/h, Distance = 216 km
\( \text{Time taken} = \frac{216}{x} \text{ hours} \)

(ii) Speed of train = \( (x + 16) \) km/h, Distance = 208 km
\( \text{Time taken} = \frac{208}{x + 16} \text{ hours} \)

(iii) Since the train takes 2 hours less:
\( \frac{216}{x} - \frac{208}{x + 16} = 2 \)
\( \Rightarrow \frac{216(x + 16) - 208x}{x(x + 16)} = 2 \)
\( \Rightarrow 216x - 208x + 3456 = 2x(x + 16) \)
\( \Rightarrow 8x + 3456 = 2x^2 + 32x \)
\( \Rightarrow 2x^2 + 32x - 8x - 3456 = 0 \)
\( \Rightarrow 2x^2 + 24x - 3456 = 0 \)
\( \Rightarrow x^2 + 12x - 1728 = 0 \)
\( \Rightarrow x^2 + 48x - 36x - 1728 = 0 \)
\( \Rightarrow x(x + 48) - 36(x + 48) = 0 \)
\( \Rightarrow (x - 36)(x + 48) = 0 \)
\( \Rightarrow x = 36 \text{ or } x = -48 \)
Since speed cannot be negative, \( x \neq -48 \). Therefore, \( x = 36 \) km/h.

Equation: \( \frac{216}{x} - \frac{208}{x + 16} = 2 \)

(iv) Speed of train = \( x + 16 = 36 + 16 = 52 \) km/h
In simple words: The car goes 216 km in 6 hours (at 36 km/h). The train goes 208 km in 4 hours (at 52 km/h). The train is 2 hours faster.

Exam Tip: In part (iv), after finding x, always substitute back into the speed expression given in the problem statement to find the final answer.

Question 31. An aeroplane flying with a wind of 30 km/h takes 40 minutes less to fly 3600 km than what it would have taken to fly against the same wind. Find the plane's speed of flying in still air.
Answer: Let the speed of the plane in still air be \( x \) km/h. Speed of wind = 30 km/h. Speed with wind = \( (x + 30) \) km/h, Speed against wind = \( (x - 30) \) km/h. Note that 40 minutes = \( \frac{40}{60} = \frac{2}{3} \) hours. Since flying with the wind takes \( \frac{2}{3} \) hours less:
\( \frac{3600}{x - 30} - \frac{3600}{x + 30} = \frac{2}{3} \)
\( \Rightarrow \frac{3600(x + 30) - 3600(x - 30)}{(x - 30)(x + 30)} = \frac{2}{3} \)
\( \Rightarrow \frac{3600x + 108000 - 3600x + 108000}{x^2 - 900} = \frac{2}{3} \)
\( \Rightarrow \frac{216000}{x^2 - 900} = \frac{2}{3} \)
\( \Rightarrow 216000 \times 3 = 2(x^2 - 900) \)
\( \Rightarrow 648000 = 2x^2 - 1800 \)
\( \Rightarrow 324000 = x^2 - 900 \)
\( \Rightarrow x^2 - 900 - 324000 = 0 \)
\( \Rightarrow x^2 - 324900 = 0 \)
\( \Rightarrow x^2 - (570)^2 = 0 \)
\( \Rightarrow (x - 570)(x + 570) = 0 \)
\( \Rightarrow x = 570 \text{ or } x = -570 \)
Since speed cannot be negative, \( x \neq -570 \). Therefore, the speed of the aeroplane in still air is 570 km/h.
In simple words: Flying with a 30 km/h tailwind, the plane goes at 600 km/h and covers 3600 km in 6 hours. Against the wind at 540 km/h, it takes 6.67 hours - which is 40 minutes longer.

Exam Tip: In wind/current problems, remember: with wind (or current), speeds add; against wind (or current), speeds subtract. The faster journey always takes less time.

Question 32. A school bus transported an excursion party to a picnic spot 150 km away. While returning, it was raining and the bus had to reduce its speed by 5 km/h, and it took one hour longer to make the return trip. Find the time taken to return.
Answer: Let the original speed of the bus be \( x \) km/h. The reduced speed during return is \( (x - 5) \) km/h. Since the return journey takes 1 hour longer:
\( \frac{150}{x - 5} - \frac{150}{x} = 1 \)
\( \Rightarrow \frac{150x - 150(x - 5)}{x(x - 5)} = 1 \)
\( \Rightarrow \frac{150x - 150x + 750}{x(x - 5)} = 1 \)
\( \Rightarrow \frac{750}{x(x - 5)} = 1 \)
\( \Rightarrow 750 = x(x - 5) \)
\( \Rightarrow x^2 - 5x - 750 = 0 \)
\( \Rightarrow x^2 - 30x + 25x - 750 = 0 \)
\( \Rightarrow x(x - 30) + 25(x - 30) = 0 \)
\( \Rightarrow (x + 25)(x - 30) = 0 \)
\( \Rightarrow x = 30 \text{ or } x = -25 \)
Since speed cannot be negative, \( x \neq -25 \). Therefore, the original speed is 30 km/h, and the return speed is \( 30 - 5 = 25 \) km/h. Time taken to return = \( \frac{150}{25} = 6 \) hours.
In simple words: On the way to the picnic at 30 km/h, the bus took 5 hours. On the return at 25 km/h, it took 6 hours - which is 1 hour longer due to the rain.

Exam Tip: The question asks for the time taken to return (not the speed) - make sure you find the final answer by dividing the distance by the reduced speed: Time = 150 ÷ 25 = 6 hours.

Question 31. Let the speed of bus while reaching picnic spot be x km/h. Since the speed of bus decreases by 5 km/h due to rain, the speed of bus in rain is (x - 5) km/h. Time taken to reach picnic spot is \( \frac{150}{x} \) and time taken to reach back to school is \( \frac{150}{x-5} \). According to the given condition, \( \frac{150}{x-5} - \frac{150}{x} = 1 \). Solve for x.

Exam Tip: Check that the speed value is positive and makes sense in the problem context - a negative speed has no physical meaning and must be rejected.

Question 32. A boat can cover 10 km up the stream and 5 km down the stream in 6 hours. If the speed of the stream is 1.5 km/h, find the speed of the boat in still water.
Answer: Let the speed of the boat in still water be x km/h. The stream moves at 1.5 km/h, so the boat's speed going upstream is (x - 1.5) km/h and downstream is (x + 1.5) km/h. The time for the upstream journey equals \( \frac{10}{x-1.5} \) hours, and the time for the downstream journey equals \( \frac{5}{x+1.5} \) hours. Since the total time is 6 hours:

\( \frac{10}{x-1.5} + \frac{5}{x+1.5} = 6 \)

Multiplying through by the common denominator \( (x-1.5)(x+1.5) \):

\( 10(x+1.5) + 5(x-1.5) = 6(x^2-2.25) \)

\( 10x + 15 + 5x - 7.5 = 6x^2 - 13.5 \)

\( 15x + 7.5 = 6x^2 - 13.5 \)

\( 6x^2 - 15x - 21 = 0 \)

\( 2x^2 - 5x - 7 = 0 \)

\( (x+1)(2x-7) = 0 \)

This gives x = -1 or x = 3.5. Since speed cannot be negative, x = 3.5 km/h.
In simple words: The boat moves faster going with the current than against it. We set up an equation showing that the total time (upstream time plus downstream time) equals 6 hours, then solve to find the boat's speed in still water is 3.5 km/h.

Exam Tip: Always reject negative solutions for speed or distance problems, as these quantities must be positive in real situations.

Question 33. Two pipes running together can fill a tank in \( 11\frac{1}{9} \) minutes. If one pipe takes 5 minutes more than the other to fill the tank, find the time in which each pipe would fill the tank.
Answer: Both pipes together fill the tank in \( 11\frac{1}{9} \) minutes, which is \( \frac{100}{9} \) minutes. So in one minute, the two pipes fill \( \frac{9}{100} \) of the tank. Let the first pipe fill the tank alone in x minutes and the second in (x + 5) minutes. In one minute, the first pipe fills \( \frac{1}{x} \) of the tank and the second fills \( \frac{1}{x+5} \) of the tank. Together:

\( \frac{1}{x} + \frac{1}{x+5} = \frac{9}{100} \)

\( \frac{x+5+x}{x(x+5)} = \frac{9}{100} \)

\( \frac{2x+5}{x^2+5x} = \frac{9}{100} \)

\( 100(2x+5) = 9(x^2+5x) \)

\( 200x + 500 = 9x^2 + 45x \)

\( 9x^2 - 155x - 500 = 0 \)

\( 9x^2 - 180x + 25x - 500 = 0 \)

\( 9x(x-20) + 25(x-20) = 0 \)

\( (9x+25)(x-20) = 0 \)

This gives x = 20 or x = -\( \frac{25}{9} \). Since time must be positive, x = 20 minutes. The first pipe fills the tank in 20 minutes and the second in 25 minutes.
In simple words: When two pipes work together, their rates of work add up. We used the combined rate to find when each pipe alone would fill the tank completely.

Exam Tip: In work-rate problems, always express rates as fractions of the job completed per unit time, then add the individual rates to get the combined rate.

Question 34. Rs.480 is divided equally among 'x' children. If the number of children were 20 more, then each would have got Rs.12 less. Find 'x'.
Answer: The total amount is Rs.480, shared equally among x children, so each child receives \( \frac{480}{x} \) rupees. If there were 20 more children, each would get \( \frac{480}{x+20} \) rupees. Since each child would get Rs.12 less:

\( \frac{480}{x} - \frac{480}{x+20} = 12 \)

Dividing the entire equation by 12:

\( \frac{40}{x} - \frac{40}{x+20} = 1 \)

\( \frac{40(x+20) - 40x}{x(x+20)} = 1 \)

\( \frac{40x + 800 - 40x}{x^2+20x} = 1 \)

\( \frac{800}{x^2+20x} = 1 \)

\( x^2 + 20x = 800 \)

\( x^2 + 20x - 800 = 0 \)

\( x^2 + 40x - 20x - 800 = 0 \)

\( x(x+40) - 20(x+40) = 0 \)

\( (x-20)(x+40) = 0 \)

This gives x = 20 or x = -40. Since the number of children cannot be negative, x = 20 children.
In simple words: Each child gets less money when more children share the same total. We set up an equation showing the difference in amounts for the two cases and solved for the original number of children.

Exam Tip: In distribution problems, check that your final answer makes sense - the number of people or items must be a positive whole number.

Question 35. Rs.7500 were divided equally among a certain number of children. Had there been 20 less children, each would have received Rs.100 more. Find the original number of children.
Answer: Rs.7500 is shared equally among x children, so each receives \( \frac{7500}{x} \) rupees. If there were 20 fewer children, each would get \( \frac{7500}{x-20} \) rupees. Since each would get Rs.100 more:

\( \frac{7500}{x-20} - \frac{7500}{x} = 100 \)

Dividing by 100:

\( \frac{75}{x-20} - \frac{75}{x} = 1 \)

\( \frac{75x - 75(x-20)}{x(x-20)} = 1 \)

\( \frac{75x - 75x + 1500}{x(x-20)} = 1 \)

\( \frac{1500}{x(x-20)} = 1 \)

\( 1500 = x^2 - 20x \)

\( x^2 - 20x - 1500 = 0 \)

\( x^2 - 50x + 30x - 1500 = 0 \)

\( x(x-50) + 30(x-50) = 0 \)

\( (x+30)(x-50) = 0 \)

This gives x = -30 or x = 50. Since the number of children cannot be negative, x = 50 children.
In simple words: When there are fewer people sharing a fixed total, each person gets more. We showed this difference in amounts equals Rs.100 and solved to find the original group size.

Exam Tip: Carefully identify whether you are subtracting or adding the known difference - check the wording to see if the amount increases or decreases in the new scenario.

Question 36. 2x articles cost Rs.(5x + 54) and (x + 2) similar articles cost Rs.(10x - 4); find x.
Answer: If 2x articles cost Rs.(5x + 54), the cost of each article is \( \frac{5x+54}{2x} \) rupees. If (x + 2) articles cost Rs.(10x - 4), the cost of each article is \( \frac{10x-4}{x+2} \) rupees. Since the articles are similar, their costs per unit must be equal:

\( \frac{5x+54}{2x} = \frac{10x-4}{x+2} \)

(5x + 54)(x + 2) = 2x(10x - 4)

\( 5x^2 + 10x + 54x + 108 = 20x^2 - 8x \)

\( 5x^2 + 64x + 108 = 20x^2 - 8x \)

\( 5x^2 - 20x^2 + 64x + 8x + 108 = 0 \)

\( -15x^2 + 72x + 108 = 0 \)

\( 15x^2 - 72x - 108 = 0 \)

\( 3(5x^2 - 24x - 36) = 0 \)

\( 5x^2 - 24x - 36 = 0 \)

\( 5x^2 - 30x + 6x - 36 = 0 \)

\( 5x(x-6) + 6(x-6) = 0 \)

\( (5x+6)(x-6) = 0 \)

This gives x = -\( \frac{6}{5} \) or x = 6. Since x must be positive (it represents a count), and must give whole number quantities, x = 6.
In simple words: If two groups of items are the same, then the cost per item in each group must be the same. We set these costs equal and solved for x.

Exam Tip: When dealing with quantities and costs, ensure your final answer makes physical sense - reject solutions that produce fractional or negative quantities.

Question 37. A trader buys x articles for a total cost of Rs.600.
(i) Write down the cost of one article in terms of x.
(ii) If the cost per article were Rs.5 more, the number of articles that can be bought for Rs.600 would be four less. Write down the equation in x and solve it to find x.
Answer:
(i) If x articles cost Rs.600, then the cost of one article is \( \frac{600}{x} \) rupees.

(ii) If the cost per article increases by Rs.5, the new cost per article becomes \( \frac{600}{x} + 5 \) rupees. At this higher price, the number of articles that can be bought for Rs.600 is \( \frac{600}{\frac{600}{x}+5} = \frac{600x}{600+5x} \). This quantity is four less than the original number:

\( \frac{600}{x-4} - \frac{600}{x} = 5 \)

\( \frac{600x - 600(x-4)}{x(x-4)} = 5 \)

\( \frac{600x - 600x + 2400}{x(x-4)} = 5 \)

\( \frac{2400}{x(x-4)} = 5 \)

\( 2400 = 5x(x-4) \)

\( 5x^2 - 20x = 2400 \)

\( 5x^2 - 20x - 2400 = 0 \)

\( x^2 - 4x - 480 = 0 \)

\( x^2 - 24x + 20x - 480 = 0 \)

\( x(x-24) + 20(x-24) = 0 \)

\( (x-24)(x+20) = 0 \)

This gives x = 24 or x = -20. Since the number of articles cannot be negative, x = 24 articles.
In simple words: When prices go up, you can buy fewer items with the same money. We found how many articles the trader originally bought by setting up an equation that shows this relationship.

Exam Tip: In part (i), express the cost per unit as a simple fraction. In part (ii), set up the equation carefully, showing that the difference in quantities equals the given value.

Question 38. A shopkeeper buys a certain number of books for Rs.960. If the cost per book was Rs.8 less, the number of books that could be bought for Rs.960 would be 4 more. Taking the original cost of each book to be Rs.x, write an equation in x and solve it to find the original cost of each book.
Answer: The shopkeeper spends Rs.960 on books costing Rs.x each, so the number of books bought is \( \frac{960}{x} \). If the cost per book were Rs.8 less (i.e., Rs.(x - 8) each), the shopkeeper could buy \( \frac{960}{x-8} \) books. Since this quantity is 4 more:

\( \frac{960}{x-8} - \frac{960}{x} = 4 \)

\( \frac{960x - 960(x-8)}{x(x-8)} = 4 \)

\( \frac{960x - 960x + 7680}{x(x-8)} = 4 \)

\( \frac{7680}{x(x-8)} = 4 \)

\( 7680 = 4x(x-8) \)

\( 4x^2 - 32x = 7680 \)

\( 4x^2 - 32x - 7680 = 0 \)

\( x^2 - 8x - 1920 = 0 \)

\( x^2 - 48x + 40x - 1920 = 0 \)

\( x(x-48) + 40(x-48) = 0 \)

\( (x-48)(x+40) = 0 \)

This gives x = 48 or x = -40. Since the cost cannot be negative, x = Rs.48 per book.
In simple words: Lower prices mean more items can be bought with the same money. We created an equation showing that 4 extra books can be bought when the price drops by Rs.8, and solved for the original price per book.

Exam Tip: Always check that your solution is positive and reasonable in the context of the problem before accepting it as final.

Question 39. A piece of cloth cost Rs.300. If the piece was 5 metre longer and each metre of cloth cost Rs.2 less, the cost of the piece would have remained unchanged. How long is the original piece of cloth and what is the rate per metre?
Answer: Let the original length be x metres. Since the total cost is Rs.300, the cost per metre is \( \frac{300}{x} \) rupees. In the new situation, the length is (x + 5) metres and the cost per metre is \( (\frac{300}{x} - 2) \) rupees. Since the total cost remains Rs.300:

\( (x+5) \times (\frac{300}{x} - 2) = 300 \)

\( (x+5) \times \frac{300-2x}{x} = 300 \)

\( (x+5)(300-2x) = 300x \)

\( 300x - 2x^2 + 1500 - 10x = 300x \)

\( -2x^2 + 1500 - 10x = 0 \)

\( -2(x^2 + 5x - 750) = 0 \)

\( x^2 + 5x - 750 = 0 \)

\( x^2 + 30x - 25x - 750 = 0 \)

\( x(x+30) - 25(x+30) = 0 \)

\( (x+30)(x-25) = 0 \)

This gives x = -30 or x = 25. Since length cannot be negative, x = 25 metres. The cost per metre is \( \frac{300}{25} = 12 \) rupees per metre.
In simple words: A longer piece at a cheaper rate per metre can cost the same as the original. We set up an equation showing that the new total (longer length times lower cost per metre) equals the original total cost.

Exam Tip: When total cost stays constant, longer length must be paired with lower unit cost. Verify your answer by checking that the new scenario actually gives the same total cost.

Question 40. The hotel bill for a number of people for overnight stay is Rs.4800. If there were 4 more the bill each person had to pay would have reduced by Rs.200. Find the number of people staying overnight.
Answer: The total bill is Rs.4800 shared equally among x people, so each pays \( \frac{4800}{x} \) rupees. If there were 4 more people, each would pay \( \frac{4800}{x+4} \) rupees. Since the bill per person reduces by Rs.200:

\( \frac{4800}{x} - \frac{4800}{x+4} = 200 \)

Dividing by 200:

\( \frac{24}{x} - \frac{24}{x+4} = 1 \)

\( \frac{24(x+4) - 24x}{x(x+4)} = 1 \)

\( \frac{24x + 96 - 24x}{x(x+4)} = 1 \)

\( \frac{96}{x(x+4)} = 1 \)

\( 96 = x^2 + 4x \)

\( x^2 + 4x - 96 = 0 \)

\( x^2 + 12x - 8x - 96 = 0 \)

\( x(x+12) - 8(x+12) = 0 \)

\( (x-8)(x+12) = 0 \)

This gives x = 8 or x = -12. Since the number of people cannot be negative, x = 8 people.
In simple words: When more people share the same bill, each person pays less. We found how many people originally stayed by showing that 4 extra people reduces each person's share by Rs.200.

Exam Tip: In cost-sharing problems, always check your answer by calculating the per-person amounts in both scenarios to verify the difference matches what was stated in the problem.

Question 41. A person was given Rs.3000 for a tour. If he extends his tour programme by 5 days, he must cut down his daily expense by Rs.20. Find the number of days of his tour programme.
Answer: Let the number of days planned for the trip be x. The total budget is Rs.3000, so the daily expense becomes Rs. 3000/x. When the trip is extended by 5 days, the total days become (x + 5). The daily expense must then be cut by Rs.20, making it Rs. (3000/x - 20). Since the total amount remains the same:

\( (x + 5)\left(\frac{3000}{x} - 20\right) = 3000 \)

Expanding and simplifying:

\( (x + 5)(3000 - 20x) = 3000x \)

\( 3000x - 20x^2 + 15000 - 100x = 3000x \)

\( -20x^2 - 100x + 15000 = 0 \)

\( x^2 + 5x - 750 = 0 \)

\( x^2 + 30x - 25x - 750 = 0 \)

\( x(x + 30) - 25(x + 30) = 0 \)

\( (x - 25)(x + 30) = 0 \)

So x = 25 or x = -30. Since the number of days cannot be negative, x = 25. The tour programme lasts 25 days.
In simple words: Set up the equation using the fact that daily expense times the number of days equals the total amount. Solve the resulting quadratic to find the number of days.

Exam Tip: Always check that your solution makes sense in context - here, negative days must be rejected. Use the condition that total amount spent remains constant to form the equation.

Question 42. The sum of the ages of Vivek and his younger brother Amit is 47 years. The product of their ages in years is 550. Find their ages.
Answer: Let Vivek's age be x years. Since the sum of their ages is 47 years, Amit's age is (47 - x) years. The product of their ages equals 550:

\( x(47 - x) = 550 \)

\( 47x - x^2 = 550 \)

\( x^2 - 47x + 550 = 0 \)

\( x^2 - 25x - 22x + 550 = 0 \)

\( x(x - 25) - 22(x - 25) = 0 \)

\( (x - 22)(x - 25) = 0 \)

So x = 22 or x = 25. If Vivek is 25 years old, Amit is 22 years old. If Vivek is 22 years old, Amit is 25 years old. Since Amit is the younger brother, Vivek's age is 25 years and Amit's age is 22 years.
In simple words: Form a quadratic equation using the sum and product conditions. Solve it to get two ages, then use the "younger brother" condition to identify which age belongs to whom.

Exam Tip: Always use all given conditions (here, "younger brother") to select the correct answer from multiple solutions. Verify by checking both the sum and product.

Question 43. Paul is x years old and his father's age is twice the square of Paul's age. Ten years hence, father's age will be four times Paul's age. Find their present ages.
Answer: Let Paul's present age be x years. His father's age is 2x² years. In 10 years, Paul will be (x + 10) years old and his father will be (2x² + 10) years old. According to the problem, the father's age after 10 years will be four times Paul's age after 10 years:

\( 2x^2 + 10 = 4(x + 10) \)

\( 2x^2 + 10 = 4x + 40 \)

\( 2x^2 - 4x - 30 = 0 \)

\( x^2 - 2x - 15 = 0 \)

\( x^2 - 5x + 3x - 15 = 0 \)

\( x(x - 5) + 3(x - 5) = 0 \)

\( (x + 3)(x - 5) = 0 \)

So x = -3 or x = 5. Since age cannot be negative, x = 5. Therefore, Paul's present age is 5 years and his father's present age is 2(5)² = 50 years.
In simple words: Use the relationship between their ages now and in the future to form a quadratic. Reject negative solutions since age must be positive.

Exam Tip: Always set up equations based on future conditions carefully. Here, "father's age will be four times Paul's age" applies to ages 10 years from now, not present ages.

Question 44. The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than 3 times the age of his son. Find their present ages.
Answer: Let the son's present age be x years. The man's age is 2x² years. After 8 years, the son will be (x + 8) years old and the man will be (2x² + 8) years old. According to the condition, the man's age after 8 years will be 4 years more than 3 times the son's age after 8 years:

\( 2x^2 + 8 = 3(x + 8) + 4 \)

\( 2x^2 + 8 = 3x + 24 + 4 \)

\( 2x^2 + 8 = 3x + 28 \)

\( 2x^2 - 3x - 20 = 0 \)

\( 2x^2 - 8x + 5x - 20 = 0 \)

\( 2x(x - 4) + 5(x - 4) = 0 \)

\( (x - 4)(2x + 5) = 0 \)

So x = 4 or x = -5/2. Since age cannot be negative or fractional, x = 4. Therefore, the son's present age is 4 years and the man's present age is 2(4)² = 32 years.
In simple words: Write an equation based on the future age relationship. The phrase "4 years more than 3 times" means add 4 to the triple of the age.

Exam Tip: Pay close attention to phrases like "4 years more than 3 times" - this translates to 3x + 4, not 3(x + 4). Always verify your answer by substituting back into both original conditions.

Question 45. Two years ago, a man's age was three times the square of his daughter's age. Three years hence, his age will be four times his daughter's age. Find their present ages.
Answer: Let the daughter's present age be x years. Two years ago, she was (x - 2) years old, and the man's age then was 3(x - 2)² years. So the man's present age is 3(x - 2)² + 2 years. Three years later, the daughter will be (x + 3) years old and the man will be 3(x - 2)² + 2 + 3 years old. According to the condition:

\( 3(x - 2)^2 + 2 + 3 = 4(x + 3) \)

\( 3(x^2 - 4x + 4) + 5 = 4x + 12 \)

\( 3x^2 - 12x + 12 + 5 = 4x + 12 \)

\( 3x^2 - 12x - 4x + 17 - 12 = 0 \)

\( 3x^2 - 16x + 5 = 0 \)

\( 3x^2 - 15x - x + 5 = 0 \)

\( 3x(x - 5) - 1(x - 5) = 0 \)

\( (3x - 1)(x - 5) = 0 \)

So x = 1/3 or x = 5. Since age cannot be a fraction, x = 5. The daughter's present age is 5 years. The man's present age is 3(5 - 2)² + 2 = 3(9) + 2 = 29 years.
In simple words: Work with ages from two different time points - past and future. Express both in terms of the daughter's present age, then set up an equation using the future condition.

Exam Tip: When dealing with past and future ages, carefully track what each expression represents. Use the condition given for the future to form the main equation, then solve and verify.

Question 46. The length (in cm) of the hypotenuse of a right angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of other side by 1 cm. Find the length of each side. Also find the perimeter and the area of the triangle.
Answer: Let the two sides (other than the hypotenuse) be x cm and y cm. From the given conditions:

Hypotenuse = x + 2 (in terms of the first side)

Hypotenuse = 2y + 1 (in terms of the second side)

Since both equal the hypotenuse:

\( x + 2 = 2y + 1 \)

\( x = 2y - 1 \)

Using the Pythagorean theorem: \( x^2 + y^2 = (2y + 1)^2 \)

Substituting x = 2y - 1:

\( (2y - 1)^2 + y^2 = (2y + 1)^2 \)

\( 4y^2 - 4y + 1 + y^2 = 4y^2 + 4y + 1 \)

\( 5y^2 - 4y + 1 = 4y^2 + 4y + 1 \)

\( y^2 - 8y = 0 \)

\( y(y - 8) = 0 \)

So y = 0 or y = 8. Since length cannot be zero, y = 8. Therefore x = 2(8) - 1 = 15 and the hypotenuse = 2(8) + 1 = 17. Perimeter = 8 + 15 + 17 = 40 cm. Area = \( \frac{1}{2} \times 8 \times 15 = 60 \) cm².
In simple words: Set up two expressions for the hypotenuse from the given conditions, equate them to find a relationship between the two sides, then use the Pythagorean theorem to solve.

Exam Tip: Always verify using the Pythagorean theorem: 8² + 15² should equal 17² (64 + 225 = 289). Don't forget to find the perimeter and area as asked in the question.

Question 47. If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 cm². However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm². Determine the sides of two squares.
Answer: Let the sides of the smaller and larger squares be x cm and y cm respectively. The area of the smaller square is x² cm² and the area of the larger square is y² cm². From the given conditions:

\( y^2 - 2x^2 = 14 \) ... (Equation 1)

\( 2y^2 + 3x^2 = 203 \) ... (Equation 2)

From Equation 1: \( y^2 = 14 + 2x^2 \)

Substituting into Equation 2:

\( 2(14 + 2x^2) + 3x^2 = 203 \)

\( 28 + 4x^2 + 3x^2 = 203 \)

\( 7x^2 = 175 \)

\( x^2 = 25 \)

\( x = 5 \) (taking the positive root)

From Equation 1: \( y^2 = 14 + 2(25) = 64 \)

\( y = 8 \)

The side of the larger square is 8 cm and the side of the smaller square is 5 cm.
In simple words: Write two equations from the two conditions. Use substitution to eliminate one variable, then solve the resulting equation to find both side lengths.

Exam Tip: When you get x² = 25, remember to take only the positive square root since length is always positive. Always verify both solutions by substituting back into the original conditions.

Multiple Choice Questions

Question 1. Which of the following is not a quadratic equation?
(1) (x + 2)² = 2(x + 3)
(2) x² + 3x = (-1)(1 - 3x)
(3) (x + 2)(x - 1) = x² - 2x - 3
(4) x³ - x² + 2x + 1 = (x + 1)³
Answer: (3) (x + 2)(x - 1) = x² - 2x - 3
In simple words: A quadratic equation has x² as its highest power, not higher. Option 3 simplifies to a linear equation (3x + 1 = 0), so it is not quadratic. All others reduce to quadratic form.

Exam Tip: Expand and simplify each option fully to see what the highest power of x becomes after cancellation. A quadratic equation must have degree 2 when written in standard form.

Question 2. If 3 is a root of the quadratic equation x² - px + 3 = 0 then the value of p is:
(1) 4
(2) 3
(3) 5
(4) 2
Answer: (1) 4
In simple words: If 3 is a root, substitute x = 3 into the equation. This gives 9 - 3p + 3 = 0, which simplifies to 3p = 12, so p = 4.

Exam Tip: Always substitute the given root into the equation directly - this is the quickest way to find unknown coefficients. No need to solve the entire equation.

Question 3. The roots of the equation x² - 3x - 10 = 0 are
(1) 2, -5
(2) -2, 5
(3) 2, 5
(4) -2, -5
Answer: (2) -2, 5
In simple words: Factor the equation as (x + 2)(x - 5) = 0. This gives x = -2 or x = 5. You can verify by checking that (-2)² - 3(-2) - 10 = 4 + 6 - 10 = 0 and (5)² - 3(5) - 10 = 25 - 15 - 10 = 0.

Exam Tip: After finding the roots by factoring, always verify by substituting back into the original equation to confirm your answer is correct.

Question 4. If one root of a quadratic equation with rational coefficients is \( \frac{3 - \sqrt{5}}{2} \), then the other root is
(1) \( \frac{-3 - \sqrt{5}}{2} \)
(2) \( \frac{-3 + \sqrt{5}}{2} \)
(3) \( \frac{3 + \sqrt{5}}{2} \)
(4) \( \frac{\sqrt{3} + 5}{2} \)
Answer: (3) \( \frac{3 + \sqrt{5}}{2} \)
In simple words: When a quadratic equation has rational coefficients and one root contains a square root, the other root must be its conjugate pair - change only the sign before the square root.

Exam Tip: Remember the conjugate pair rule for irrational roots - if one root is \( a + \sqrt{b} \), the other is \( a - \sqrt{b} \).

Question 5. If the equation \( 2x^2 - 5x + (k + 3) = 0 \) has equal roots then the value of k is
(1) \( \frac{9}{8} \)
(2) \( -\frac{9}{8} \)
(3) \( \frac{1}{8} \)
(4) \( -\frac{1}{8} \)
Answer: (1) \( \frac{9}{8} \)
In simple words: For equal roots, the discriminant must equal zero. Use the formula \( b^2 - 4ac = 0 \) and solve for k.

Exam Tip: Equal roots occur when the discriminant = 0 - memorise this condition as it appears frequently in different question types.

Question 6. The value(s) of k for which the quadratic equation \( 2x^2 - kx + k = 0 \) has equal roots is (are)
(1) 0 only
(2) 4
(3) 8 only
(4) 0, 8
Answer: (4) 0, 8
In simple words: Set the discriminant equal to zero: \( k^2 - 8k = 0 \). Factor to get \( k(k - 8) = 0 \), giving two values: k = 0 or k = 8.

Exam Tip: When a quadratic equation in k results from the discriminant condition, check if there are two solutions by factoring completely.

Question 7. If the equation \( 3x^2 - kx + 2k = 0 \) has equal roots, then the value(s) of k is (are)
(1) 6
(2) 0 only
(3) 24 only
(4) 0 or 24
Answer: (4) 0 or 24
In simple words: Apply the discriminant condition: \( b^2 - 4ac = 0 \) becomes \( k^2 - 24k = 0 \). This factors as \( k(k - 24) = 0 \), so k = 0 or k = 24.

Exam Tip: Always factorise the resulting discriminant equation completely - it may yield two distinct values rather than just one.

Question 8. If the equation \( (k + 1)x^2 - 2(k - 1)x + 1 = 0 \) has equal roots, then the values of k are
(1) 1, 3
(2) 0, 3
(3) 0, 1
(4) \( 0, \frac{3}{4} \)
Answer: (2) 0, 3
In simple words: Using the discriminant condition, the equation simplifies to \( 4k^2 - 12k = 0 \). Factor as \( 4k(k - 3) = 0 \) to get k = 0 or k = 3.

Exam Tip: Be careful expanding squared binomials - \( [−2(k−1)]^2 = 4(k−1)^2 \) expands fully to \( 4k^2 + 4 − 8k \).

Question 9. If the equation \( 2x^2 - 6x + p = 0 \) has real and different roots, then the values of p are given by
(1) \( p < \frac{9}{2} \)
(2) \( p \leq \frac{9}{2} \)
(3) \( p > \frac{9}{2} \)
(4) \( p \geq \frac{9}{2} \)
Answer: (1) \( p < \frac{9}{2} \)
In simple words: For real and different roots, the discriminant must be strictly greater than zero. The condition \( b^2 - 4ac > 0 \) gives \( 36 - 8p > 0 \), so \( p < \frac{9}{2} \).

Exam Tip: Real and different roots require discriminant > 0 (strict inequality), not ≥ 0 - this is a common distinction tested.

Question 10. The quadratic equation \( 2x^2 - \sqrt{5}x + 1 = 0 \) has
(1) two distinct real roots
(2) two equal real roots
(3) no real roots
(4) more than two real roots
Answer: (3) no real roots
In simple words: Calculate the discriminant: \( b^2 - 4ac = (-\sqrt{5})^2 - 4(2)(1) = 5 - 8 = -3 \). Since this is negative, the equation has no real roots.

Exam Tip: A negative discriminant immediately tells you there are no real roots - no need to attempt further solving.

Question 11. If the roots of equation \( x^2 - 6x + k = 0 \) are real and distinct, then value of k is :
(1) > - 9
(2) > - 6
(3) < 6
(4) < 9
Answer: (4) < 9
In simple words: For real and distinct roots, \( b^2 - 4ac > 0 \). This gives \( 36 - 4k > 0 \), which simplifies to \( k < 9 \).

Exam Tip: When solving an inequality from the discriminant condition, remember to reverse the inequality sign when dividing by a negative number.

Question 12. The roots of the quadratic equation \( px^2 - qx + r = 0 \) are real and equal if :
(a) \( p^2 = 4qr \)
(b) \( q^2 = 4pr \)
(c) \( -q^2 = 4pr \)
(d) \( p^2 > 4qr \)
Answer: (b) \( q^2 = 4pr \)
In simple words: For a quadratic equation with real and equal roots, the discriminant must equal zero. Using the formula \( D = b^2 - 4ac = 0 \), we get \( q^2 - 4pr = 0 \), which means \( q^2 = 4pr \).

Exam Tip: This is the key formula for equal roots with generic coefficients p, q, r - knowing which variable appears squared in the discriminant condition is essential.

Question 13. If \( x^2 + kx + 6 = (x - 2)(x - 3) \) for all values of x, then the value of k is :
(1) - 5
(2) - 3
(3) - 2
(4) 5
Answer: (1) - 5
In simple words: Expand the right side: \( (x - 2)(x - 3) = x^2 - 5x + 6 \). Comparing with the left side \( x^2 + kx + 6 \), the coefficient of x must match, so k = - 5.

Exam Tip: When two expressions are equal for all values of a variable, match the coefficients of like terms systematically.

Question 14. The roots of quadratic equation \( x^2 - 1 = 0 \) are :
(1) 0
(2) 1
(3) - 1
(4) ± 1
Answer: (4) ± 1
In simple words: Factor the equation: \( (x + 1)(x - 1) = 0 \), giving x = - 1 or x = 1. Both values satisfy the equation.

Exam Tip: Difference of squares \( a^2 - b^2 = (a+b)(a-b) \) is a quick way to factorise and find both roots immediately.

Assertion-Reason Type Questions

Question. Assertion (A): Every quadratic equation \( ax^2 + bx + c = 0, a \neq 0 \), where a, b and c are all real numbers has two real roots.
Reason (R): Every quadratic equation \( ax^2 + bx + c = 0, a \neq 0 \), where a, b and c are all real numbers has two real roots if \( b^2 - 4ac \geq 0 \).
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (2) Assertion (A) is false, but Reason (R) is true.
In simple words: Not every quadratic equation with real coefficients automatically produces real roots - it depends on whether the discriminant is non-negative. Assertion (A) is too broad, but Reason (R) correctly states the actual condition.

Exam Tip: For Assertion-Reason questions, evaluate each statement independently first - Assertion (A) must always be tested against the Reason (R) for logical connection.

Question. Assertion (A): The quadratic equation \( 4x^2 + 12x + 15 = 0 \) has no real roots.
Reason (R): The quadratic equation \( ax^2 + bx + c = 0 \) has real roots iff its 'discriminant' = \( b^2 - 4ac \geq 0 \).
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: Calculate the discriminant: \( 12^2 - 4(4)(15) = 144 - 240 = -96 \). Since it is negative, the equation has no real roots. The Reason correctly explains why this happens.

Exam Tip: Always compute the discriminant numerically to verify the Assertion before selecting your answer - this guards against careless errors.

Question. Assertion (A): The equation \( 9x^2 + 6x - k = 0 \) has real roots if \( k \geq -1 \).
Reason (R): The quadratic equation \( ax^2 + bx + c = 0 \) has real roots if 'discriminant' = \( b^2 - 4ac > 0 \).
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (1) Assertion (A) is true, but Reason (R) is false.
In simple words: For the equation to have real roots, the discriminant must satisfy \( b^2 - 4ac \geq 0 \) (with equality allowed), not just be strictly greater than zero. From \( 36 + 36k \geq 0 \), we get \( k \geq -1 \), confirming Assertion (A). Reason (R) is incomplete because it requires strict inequality instead of allowing equality.

Exam Tip: Distinguish carefully between > 0 (strict inequality, for distinct roots) and ≥ 0 (non-strict inequality, for real roots including equal roots).

Question. Consider the polynomial \( 2x^2 - 3x + 5 \)
Assertion (A): Factorisation of the above polynomial is not possible.

Question. Assertion (A): The polynomial 2x² - 3x + 5 cannot be broken down into linear factors with real coefficients.
Answer: We need to find the discriminant of the given polynomial 2x² - 3x + 5. Using the formula Discriminant (D) = b² - 4ac, where a = 2, b = -3, and c = 5, we calculate: D = (-3)² - 4 × 2 × 5 = 9 - 40 = -31. Since the discriminant is negative, this quadratic equation has no real roots. When a quadratic has no real roots, it cannot be factorised into linear factors with real coefficients. Therefore, Assertion (A) is correct. The Reason (R) states that the discriminant 'b² - 4ac' is negative, which is also true and correctly explains why the assertion is true. So both the assertion and reason are correct, and the reason provides the right explanation for the assertion.
In simple words: When you work out the discriminant and get a negative number, it means the equation has no real roots and cannot be split into simpler pieces with real numbers.

Exam Tip: Always compute the discriminant first - if it's negative, the quadratic cannot have real roots and cannot be factored into linear terms with real coefficients. This is a key connection that examiners test frequently.

Question 5. Consider the following equation k²x² - 2kx + 1 = 0. Assertion (A): This equation has real roots for all non-zero values of k. Reason (R): The discriminant of this equation is zero.
Answer: Given the equation k²x² - 2kx + 1 = 0, we compare it with the standard form ax² + bx + c = 0. This gives us a = k², b = -2k, and c = 1. Now we calculate the discriminant: D = b² - 4ac = (-2k)² - 4 × k² × 1 = 4k² - 4k² = 0. Since the discriminant equals zero, the equation has one repeated real root (or equal roots). This holds true for all non-zero values of k. Therefore, Assertion (A) is correct because the equation indeed has real roots for all non-zero values of k. Reason (R) is also correct since the discriminant is always zero. Moreover, the reason correctly explains the assertion - the reason why the equation has real roots is precisely because the discriminant is zero. Thus, both statements are correct, and the reason provides the proper justification for the assertion.
In simple words: When the discriminant is zero, the quadratic equation always has real roots (two equal ones). Since the discriminant stays zero no matter what non-zero value k takes, the equation always has real roots.

Exam Tip: Remember that D = 0 means equal/repeated real roots. When testing assertion-reason pairs, always check if the reason actually explains the assertion - correct facts alone aren't enough if they don't connect logically.

Chapter Test

Question 1(i). Solve the following equations by factorisation: x² + 6x - 16 = 0
Answer: We need to break down this equation into simpler parts. Starting with x² + 6x - 16 = 0, we split the middle term: x² + 8x - 2x - 16 = 0. Now we group: x(x + 8) - 2(x + 8) = 0. Factoring out the common term (x + 8): (x - 2)(x + 8) = 0. This gives us two equations: x - 2 = 0 or x + 8 = 0. Solving each: x = 2 or x = -8. Therefore, the roots are 2 and -8.
In simple words: We split the middle term in a smart way, then group and factor. When the product equals zero, at least one factor must be zero, giving us our answers.

Exam Tip: For splitting the middle term, find two numbers that multiply to give ac (here, 1 × (-16) = -16) and add to give b (here, 6). The pair is 8 and -2. Correct factorisation is crucial - take time to verify your brackets multiply back to the original.

Question 1(ii). Solve the following equations by factorisation: 3x² + 11x + 10 = 0
Answer: We begin with 3x² + 11x + 10 = 0. We split the middle term into two parts that multiply to give 3 × 10 = 30 and add to give 11. These numbers are 6 and 5. So we rewrite: 3x² + 6x + 5x + 10 = 0. Grouping: 3x(x + 2) + 5(x + 2) = 0. Factoring out (x + 2): (x + 2)(3x + 5) = 0. This gives us: x + 2 = 0 or 3x + 5 = 0. Solving: x = -2 or x = -5/3. Thus the roots are -2 and -5/3.
In simple words: Find two numbers that multiply to 30 and add to 11 (these are 6 and 5). Use them to split the middle term, then group and factor out the common binomial.

Exam Tip: Always verify your answer by expanding the factored form back to the original equation. Also, present fractional roots in simplest form. Showing the grouping step clearly helps you avoid errors and earns method marks.

Question 2(i). Solve the following equations by factorisation: 2x² + ax - a² = 0
Answer: Given 2x² + ax - a² = 0, we look for two numbers that multiply to 2 × (-a²) = -2a² and add to a. These are 2a and -a. We rewrite the equation as: 2x² + 2ax - ax - a² = 0. Grouping the terms: 2x(x + a) - a(x + a) = 0. Factoring out the common binomial (x + a): (x + a)(2x - a) = 0. This gives us two cases: x + a = 0 or 2x - a = 0. Solving: x = -a or x = a/2. Therefore the roots are -a and a/2.
In simple words: We split the middle term using the product-sum method, group into two parts, then factor out what's common. This works even when the coefficients involve letters.

Exam Tip: When equations contain letters as coefficients (like a here), treat them as constants during factorisation. Present final answers in terms of those letters. Always double-check by substituting back if possible.

Question 2(ii). Solve the following equations by factorisation: \( \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \)
Answer: We have \( \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \). We need two numbers that multiply to \( \sqrt{3} \times 7\sqrt{3} = 21 \) and add to 10. These are 7 and 3. Rewriting: \( \sqrt{3}x^2 + 7x + 3x + 7\sqrt{3} = 0 \). Grouping: \( x(\sqrt{3}x + 7) + \sqrt{3}(\sqrt{3}x + 7) = 0 \). Factoring out \( (\sqrt{3}x + 7) \): \( (x + \sqrt{3})(\sqrt{3}x + 7) = 0 \). This gives: \( x + \sqrt{3} = 0 \) or \( \sqrt{3}x + 7 = 0 \). Solving: \( x = -\sqrt{3} \) or \( x = -\frac{7}{\sqrt{3}} = -\frac{7\sqrt{3}}{3} \). Thus the roots are \( -\sqrt{3} \) and \( -\frac{7\sqrt{3}}{3} \).
In simple words: Even with square roots in the equation, the same splitting and grouping method works. Just be careful when rationalising the denominator at the end.

Exam Tip: When surds appear, always rationalise any denominator that contains a surd. Present roots in their simplest rationalised form. The product-sum method still applies - the numbers you're looking for should still be simple integers.

Question 3(i). Solve the following equations by factorisation: x(x + 1) + (x + 2)(x + 3) = 42
Answer: Starting with x(x + 1) + (x + 2)(x + 3) = 42, we expand each part. The first product gives x² + x. The second product gives x² + 3x + 2x + 6 = x² + 5x + 6. Adding them together: x² + x + x² + 5x + 6 = 42. Combining like terms: 2x² + 6x + 6 = 42. Moving 42 to the left side: 2x² + 6x + 6 - 42 = 0, which simplifies to 2x² + 6x - 36 = 0. Dividing by 2: x² + 3x - 18 = 0. We split the middle term using numbers that multiply to -18 and add to 3, which are 6 and -3. Rewriting: x² + 6x - 3x - 18 = 0. Grouping: x(x + 6) - 3(x + 6) = 0. Factoring: (x - 3)(x + 6) = 0. Therefore: x - 3 = 0 or x + 6 = 0, giving x = 3 or x = -6. The roots are 3 and -6.
In simple words: Expand each product, add them together, simplify the equation, then factorise it like a normal quadratic. Breaking down complex-looking equations into standard form is the key.

Exam Tip: Always expand carefully and combine all like terms before attempting to factorise. Dividing through by common factors (like 2 here) makes the numbers smaller and easier to work with. Verify your roots by substituting back into the original equation.

Question 3(ii). Solve the following equations by factorisation: \( \frac{6}{x} - \frac{2}{x-1} = \frac{1}{x-2} \)
Answer: Given \( \frac{6}{x} - \frac{2}{x-1} = \frac{1}{x-2} \), we first find a common denominator for the left side. The left side becomes: \( \frac{6(x-1) - 2x}{x(x-1)} = \frac{6x - 6 - 2x}{x(x-1)} = \frac{4x - 6}{x(x-1)} = \frac{1}{x-2} \). Now we cross-multiply: (4x - 6)(x - 2) = x(x - 1). Expanding the left side: 4x² - 8x - 6x + 12 = x² - x. This gives 4x² - 14x + 12 = x² - x. Moving all terms to one side: 4x² - x² - 14x + x + 12 = 0, which simplifies to 3x² - 13x + 12 = 0. We split the middle term using numbers that multiply to 36 and add to -13, which are -9 and -4. Rewriting: 3x² - 9x - 4x + 12 = 0. Grouping: 3x(x - 3) - 4(x - 3) = 0. Factoring: (3x - 4)(x - 3) = 0. Therefore: 3x - 4 = 0 or x - 3 = 0, giving x = 4/3 or x = 3. The roots are 4/3 and 3.
In simple words: Combine fractions on one side by finding a common denominator, then cross-multiply to remove the fractions. Solve the resulting quadratic using factorisation.

Exam Tip: With fraction equations, always note any values that make denominators zero (these cannot be roots). After solving, verify that your roots don't violate these restrictions. Show all steps in combining fractions clearly.

Question 4(i). Solve the following equations by factorisation: \( \sqrt{x + 15} = x + 3 \)
Answer: Given \( \sqrt{x + 15} = x + 3 \), we square both sides to eliminate the square root: x + 15 = (x + 3)². Expanding the right side: x + 15 = x² + 6x + 9. Rearranging: x² + 6x + 9 - x - 15 = 0, which simplifies to x² + 5x - 6 = 0. We split the middle term using numbers that multiply to -6 and add to 5, which are 6 and -1. Rewriting: x² + 6x - x - 6 = 0. Grouping: x(x + 6) - 1(x + 6) = 0. Factoring: (x - 1)(x + 6) = 0. This gives x - 1 = 0 or x + 6 = 0, so x = 1 or x = -6. However, when we squared the equation, we may have introduced extraneous solutions. We must check both roots. For x = -6: \( \sqrt{-6 + 15} = -6 + 3 \) becomes \( \sqrt{9} = -3 \), or 3 = -3, which is false. For x = 1: \( \sqrt{1 + 15} = 1 + 3 \) becomes \( \sqrt{16} = 4 \), or 4 = 4, which is true. Therefore, only x = 1 is a valid root.
In simple words: Square both sides to get a quadratic, solve it by factorisation, then check both answers in the original equation because squaring can create false solutions.

Exam Tip: Whenever you square an equation (or take any other non-reversible step), always verify your solutions in the original equation. Extraneous solutions are a common source of marks lost. Show your verification work clearly.

Question 4(ii). Solve the following equations by factorisation: \( \sqrt{3x^2 - 2x - 1} = 2x - 2 \)
Answer: Given \( \sqrt{3x^2 - 2x - 1} = 2x - 2 \), we square both sides: 3x² - 2x - 1 = (2x - 2)². Expanding the right side: 3x² - 2x - 1 = 4x² + 4 - 8x. Rearranging: 3x² - 4x² - 2x + 8x - 1 - 4 = 0, which simplifies to -x² + 6x - 5 = 0. Multiplying by -1: x² - 6x + 5 = 0. We split the middle term using numbers that multiply to 5 and add to -6, which are -5 and -1. Rewriting: x² - 5x - x + 5 = 0. Grouping: x(x - 5) - 1(x - 5) = 0. Factoring: (x - 1)(x - 5) = 0. This gives x = 1 or x = 5. We must verify both in the original equation. For x = 1: \( \sqrt{3(1)^2 - 2(1) - 1} = 2(1) - 2 \) becomes \( \sqrt{3 - 2 - 1} = 0 \), or \( \sqrt{0} = 0 \), which is true. For x = 5: \( \sqrt{3(5)^2 - 2(5) - 1} = 2(5) - 2 \) becomes \( \sqrt{75 - 10 - 1} = 8 \), or \( \sqrt{64} = 8 \), which is true. Therefore, both x = 1 and x = 5 are valid roots.
In simple words: Square both sides carefully, expand and simplify to get a standard quadratic, factorise to find potential answers, then test each one in the original equation.

Exam Tip: Always check your solutions against the original equation - one or both may be extraneous. Show the substitution and verification clearly. Note that the expression under the square root must be non-negative.

Question 5(i). Solve the following equations by using formula: 2x² - 3x - 1 = 0
Answer: The given equation is 2x² - 3x - 1 = 0. Comparing with the standard form ax² + bx + c = 0, we have a = 2, b = -3, and c = -1. Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we substitute: \( x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 2 \times (-1)}}{2 \times 2} = \frac{3 \pm \sqrt{9 + 8}}{4} = \frac{3 \pm \sqrt{17}}{4} \). Therefore, the roots are \( \frac{3 + \sqrt{17}}{4} \) and \( \frac{3 - \sqrt{17}}{4} \).
In simple words: Identify a, b, and c from the equation, plug them into the quadratic formula, and simplify carefully. The formula always works even when factorisation is difficult.

Exam Tip: Be extra careful with signs when substituting into the formula, especially when b or c is negative. Write the discriminant separately and simplify it before inserting into the final answer. Always simplify square roots if possible (e.g., \( \sqrt{8} = 2\sqrt{2} \)).

Question 5(ii). Solve the following equations by using formula: \( x\left(3x + \frac{1}{2}\right) = 6 \)
Answer: The given equation is \( x\left(3x + \frac{1}{2}\right) = 6 \). Expanding: \( 3x^2 + \frac{1}{2}x = 6 \). Multiplying through by 2 to clear fractions: 6x² + x = 12. Rearranging to standard form: 6x² + x - 12 = 0. Comparing with ax² + bx + c = 0, we have a = 6, b = 1, and c = -12. Using the quadratic formula: \( x = \frac{-(1) \pm \sqrt{(1)^2 - 4 \times 6 \times (-12)}}{2 \times 6} = \frac{-1 \pm \sqrt{1 + 288}}{12} = \frac{-1 \pm \sqrt{289}}{12} = \frac{-1 \pm 17}{12} \). This gives \( x = \frac{-1 + 17}{12} = \frac{16}{12} = \frac{4}{3} \) or \( x = \frac{-1 - 17}{12} = \frac{-18}{12} = -\frac{3}{2} \). The roots are \( \frac{4}{3} \) and \( -\frac{3}{2} \).
In simple words: Expand the equation, clear any fractions by multiplying by a whole number, then use the quadratic formula. Simplify your final answers to lowest terms.

Exam Tip: Clearing fractions at the start makes arithmetic much simpler. When the discriminant is a perfect square, you get rational roots - check this before finalising. Always reduce fractions to simplest form in your answer.

Question 6(i). Solve the following equations by using formula: \( \frac{2x + 5}{3x + 4} = \frac{x + 1}{x + 3} \)
Answer: Given \( \frac{2x + 5}{3x + 4} = \frac{x + 1}{x + 3} \), we cross-multiply: (2x + 5)(x + 3) = (x + 1)(3x + 4). Expanding the left side: 2x² + 6x + 5x + 15 = 2x² + 11x + 15. Expanding the right side: 3x² + 4x + 3x + 4 = 3x² + 7x + 4. Setting them equal: 2x² + 11x + 15 = 3x² + 7x + 4. Rearranging: 2x² - 3x² + 11x - 7x + 15 - 4 = 0, which simplifies to -x² + 4x + 11 = 0. Multiplying by -1: x² - 4x - 11 = 0. Comparing with ax² + bx + c = 0, we have a = 1, b = -4, and c = -11. Using the quadratic formula: \( x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times (-11)}}{2 \times 1} = \frac{4 \pm \sqrt{16 + 44}}{2} = \frac{4 \pm \sqrt{60}}{2} = \frac{4 \pm 2\sqrt{15}}{2} = 2 \pm \sqrt{15} \). The roots are \( 2 + \sqrt{15} \) and \( 2 - \sqrt{15} \).
In simple words: Cross-multiply to turn the equation into a standard quadratic, combine all terms on one side, then apply the quadratic formula.

Exam Tip: When simplifying \( \sqrt{60} \), recognise that 60 = 4 × 15, so \( \sqrt{60} = 2\sqrt{15} \). Always simplify radicals by factoring out perfect squares. Remember that values making any denominator zero cannot be roots.

Question 6(ii). Solve the following equations by using formula: \( \frac{2}{x+2} - \frac{1}{x+1} = \frac{4}{x+4} - \frac{3}{x+3} \)
Answer: Given \( \frac{2}{x+2} - \frac{1}{x+1} = \frac{4}{x+4} - \frac{3}{x+3} \), we simplify each side separately. Left side: \( \frac{2(x+1) - (x+2)}{(x+2)(x+1)} = \frac{2x + 2 - x - 2}{(x+2)(x+1)} = \frac{x}{(x+2)(x+1)} \). Right side: \( \frac{4(x+3) - 3(x+4)}{(x+4)(x+3)} = \frac{4x + 12 - 3x - 12}{(x+4)(x+3)} = \frac{x}{(x+4)(x+3)} \). Now our equation is: \( \frac{x}{(x+2)(x+1)} = \frac{x}{(x+4)(x+3)} \). If x ≠ 0, we can divide both sides by x: \( \frac{1}{(x+2)(x+1)} = \frac{1}{(x+4)(x+3)} \). This means (x + 2)(x + 1) = (x + 4)(x + 3). Expanding: x² + 3x + 2 = x² + 7x + 12. Simplifying: 3x - 7x = 12 - 2, which gives -4x = 10, so x = -2.5 or x = -5/2. We must also check if x = 0 is a solution: substituting into the original equation gives 0 = 0, which is true. However, checking x = -5/2 in the original equation (ensuring no division by zero), we verify it works. If x = 0, the original equation is satisfied. The roots are x = 0 and x = -5/2.
In simple words: Find a common denominator on each side, simplify, then cross-multiply or compare the simplified results. Always check for special cases like x = 0 and verify no denominators become zero.

Exam Tip: When both sides simplify to the same form over different denominators, equate the denominators (only if the numerators are equal and non-zero). Always verify your solutions don't make any original denominator zero. Show all simplification steps to earn method marks.

Question 7(i). Solve the following equations by using formula: \( \frac{3x - 4}{7} + \frac{7}{3x - 4} = \frac{5}{2}, x \neq \frac{4}{3} \)
Answer: Let \( y = \frac{3x - 4}{7} \). Then \( \frac{7}{3x - 4} = \frac{1}{y} \). The equation becomes \( y + \frac{1}{y} = \frac{5}{2} \). Multiplying by y: \( y^2 + 1 = \frac{5y}{2} \). Rearranging: \( 2y^2 - 5y + 2 = 0 \). By factoring: \( (2y - 1)(y - 2) = 0 \). So \( y = \frac{1}{2} \) or \( y = 2 \).
When \( y = \frac{1}{2} \): \( \frac{3x - 4}{7} = \frac{1}{2} \) gives \( 3x - 4 = \frac{7}{2} \), so \( 3x = \frac{15}{2} \) and \( x = \frac{5}{2} \).
When \( y = 2 \): \( \frac{3x - 4}{7} = 2 \) gives \( 3x - 4 = 14 \), so \( 3x = 18 \) and \( x = 6 \).
In simple words: Replace the fraction with a single variable to make it easier. This turns it into a simple quadratic that factors nicely, giving two answers.

Exam Tip: Always check that your final answers don't make any denominator zero - both values satisfy this here.

Question 7(ii). Solve the following equations by using formula: \( \frac{x}{4} - 3 = \frac{2x + 3}{5}, x \neq 0, -\frac{3}{2} \)
Answer: Taking the common denominator of the left side: \( \frac{x - 12}{4} = \frac{2x + 3}{5} \). Cross-multiplying: \( 5(x - 12) = 4(2x + 3) \), which gives \( 5x - 60 = 8x + 12 \). Simplifying: \( -3x = 72 \), so \( x = -24 \). However, let me work from the reformulated equation in the source. The equation becomes \( \frac{4 - 3x}{x} = \frac{5}{2x + 3} \). Cross-multiplying: \( (4 - 3x)(2x + 3) = 5x \). Expanding: \( 8x + 12 - 6x^2 - 9x = 5x \). Rearranging: \( -6x^2 - 6x + 12 = 0 \) or \( 6x^2 + 6x - 12 = 0 \), which factors as \( 6(x^2 + x - 2) = 0 \). Thus \( x^2 + x - 2 = 0 \) gives \( (x + 2)(x - 1) = 0 \), so \( x = -2 \) or \( x = 1 \).
In simple words: Get all terms to one side and find a common denominator, then cross-multiply to clear fractions and solve the resulting quadratic.

Exam Tip: When equations have fractions with variables in denominators, always note the restrictions at the start and verify your solutions don't violate them.

Question 8(i). Solve the following equations by using formula: \( x^2 + (4 - 3a)x - 12a = 0 \)
Answer: Comparing with \( ax^2 + bx + c = 0 \), we have \( a = 1 \), \( b = 4 - 3a \), and \( c = -12a \). Using the quadratic formula:
\( x = \frac{-(4 - 3a) \pm \sqrt{(4 - 3a)^2 - 4(1)(-12a)}}{2} \)
\( x = \frac{3a - 4 \pm \sqrt{16 - 24a + 9a^2 + 48a}}{2} \)
\( x = \frac{3a - 4 \pm \sqrt{9a^2 + 24a + 16}}{2} \)
\( x = \frac{3a - 4 \pm \sqrt{(3a + 4)^2}}{2} \)
\( x = \frac{3a - 4 \pm (3a + 4)}{2} \)
Taking the positive sign: \( x = \frac{3a - 4 + 3a + 4}{2} = \frac{6a}{2} = 3a \)
Taking the negative sign: \( x = \frac{3a - 4 - 3a - 4}{2} = \frac{-8}{2} = -4 \)
In simple words: Plug the coefficients into the quadratic formula. The discriminant simplifies to a perfect square, making the calculation cleaner and revealing two roots.

Exam Tip: Always look for perfect square patterns in the discriminant - it saves calculation time and reduces arithmetic errors.

Question 8(ii). Solve the following equations by using formula: \( 10ax^2 - 6x + 15ax - 9 = 0, a \neq 0 \)
Answer: First, rewrite by grouping: \( 2x(5ax - 3) + 3(5ax - 3) = 0 \), giving \( (2x + 3)(5ax - 3) = 0 \). So \( x = -\frac{3}{2} \) or \( x = \frac{3}{5a} \). Alternatively, using the quadratic formula with the equation \( 10ax^2 + (15a - 6)x - 9 = 0 \), we get \( a = 10a \), \( b = 15a - 6 \), \( c = -9 \). The discriminant is:
\( \Delta = (15a - 6)^2 - 4(10a)(-9) = 225a^2 - 180a + 36 + 360a = 225a^2 + 180a + 36 = (15a + 6)^2 \)
\( x = \frac{-(15a - 6) \pm (15a + 6)}{20a} = \frac{6 - 15a \pm (15a + 6)}{20a} \)
Taking the positive sign: \( x = \frac{6 - 15a + 15a + 6}{20a} = \frac{12}{20a} = \frac{3}{5a} \)
Taking the negative sign: \( x = \frac{6 - 15a - 15a - 6}{20a} = \frac{-30a}{20a} = -\frac{3}{2} \)
In simple words: Factor out common terms from groups of monomials, or use the quadratic formula when the perfect square discriminant emerges.

Exam Tip: Factoring by grouping is often faster than the formula for this type - always scan the equation first to see if terms can be paired.

Question 9. Solve for x using the quadratic formula. Write your answer correct to two significant figures: \( (x - 1)^2 - 3x + 4 = 0 \)
Answer: Expand \( (x - 1)^2 \): \( x^2 - 2x + 1 - 3x + 4 = 0 \), giving \( x^2 - 5x + 5 = 0 \). Using the quadratic formula with \( a = 1 \), \( b = -5 \), \( c = 5 \):
\( x = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2} \)
Since \( \sqrt{5} \approx 2.236 \):
\( x = \frac{5 + 2.236}{2} = \frac{7.236}{2} \approx 3.618 \approx 3.6 \)
\( x = \frac{5 - 2.236}{2} = \frac{2.764}{2} \approx 1.382 \approx 1.4 \)
In simple words: Expand the brackets, simplify to standard form, then apply the formula and round the final answers to two significant figures as instructed.

Exam Tip: Always expand brackets fully before identifying a, b, and c - a careless mistake here can throw off the entire solution.

Question 10. Discuss the nature of the roots of the following equations. In case real roots exist, then find them.
(i) \( 3x^2 - 7x + 8 = 0 \)
(ii) \( x^2 - \frac{1}{2}x - 4 = 0 \)
(iii) \( 5x^2 - 6\sqrt{5}x + 9 = 0 \)
(iv) \( \sqrt{3}x^2 - 2x - \sqrt{3} = 0 \)
Answer:
(i) For \( 3x^2 - 7x + 8 = 0 \), the discriminant is \( \Delta = (-7)^2 - 4(3)(8) = 49 - 96 = -47 < 0 \). Since the discriminant is negative, the equation has no real roots.

(ii) For \( x^2 - \frac{1}{2}x - 4 = 0 \), the discriminant is \( \Delta = \left(-\frac{1}{2}\right)^2 - 4(1)(-4) = \frac{1}{4} + 16 = \frac{65}{4} > 0 \). The equation has two distinct real roots.
\( x = \frac{\frac{1}{2} \pm \sqrt{\frac{65}{4}}}{2} = \frac{\frac{1}{2} \pm \frac{\sqrt{65}}{2}}{2} = \frac{1 \pm \sqrt{65}}{4} \)

(iii) For \( 5x^2 - 6\sqrt{5}x + 9 = 0 \), the discriminant is \( \Delta = (-6\sqrt{5})^2 - 4(5)(9) = 180 - 180 = 0 \). The equation has two equal real roots.
\( x = \frac{6\sqrt{5} \pm 0}{10} = \frac{6\sqrt{5}}{10} = \frac{3\sqrt{5}}{5} \) (both roots are the same)

(iv) For \( \sqrt{3}x^2 - 2x - \sqrt{3} = 0 \), the discriminant is \( \Delta = (-2)^2 - 4(\sqrt{3})(-\sqrt{3}) = 4 + 12 = 16 > 0 \). The equation has two distinct real roots.
\( x = \frac{2 \pm \sqrt{16}}{2\sqrt{3}} = \frac{2 \pm 4}{2\sqrt{3}} \)
\( x = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \) or \( x = \frac{-2}{2\sqrt{3}} = -\frac{1}{\sqrt{3}} \)
In simple words: Check the discriminant first - if it is negative, no real roots exist; if zero, there is one repeated root; if positive, two different roots exist. Then find them using the formula only when roots are real.

Exam Tip: State the nature of roots before solving - examiners give marks for identifying discriminant values correctly, not just for the final answer.

Question 11. Find the values of k so that the quadratic equation \( (4 - k)x^2 + 2(k + 2)x + (8k + 1) = 0 \) has equal roots.
Answer: For equal roots, the discriminant must equal zero. With \( a = 4 - k \), \( b = 2(k + 2) = 2k + 4 \), \( c = 8k + 1 \):
\( b^2 - 4ac = 0 \)
\( (2k + 4)^2 - 4(4 - k)(8k + 1) = 0 \)
\( 4k^2 + 16k + 16 - 4(32k + 4 - 8k^2 - k) = 0 \)
\( 4k^2 + 16k + 16 - 128k - 16 + 32k^2 + 4k = 0 \)
\( 36k^2 - 108k = 0 \)
\( 36k(k - 3) = 0 \)
So \( k = 0 \) or \( k = 3 \).
In simple words: Set the discriminant equal to zero and expand carefully. Factor out common terms to find the values of k.

Exam Tip: After solving, verify by substituting back - this catches algebra errors quickly.

Question 12. Find the values of m so that the quadratic equation \( 3x^2 - 5x - 2m = 0 \) has two distinct real roots.
Answer: For two distinct real roots, the discriminant must be greater than zero. With \( a = 3 \), \( b = -5 \), \( c = -2m \):
\( b^2 - 4ac > 0 \)
\( (-5)^2 - 4(3)(-2m) > 0 \)
\( 25 + 24m > 0 \)
\( 24m > -25 \)
\( m > -\frac{25}{24} \)
In simple words: Work with the discriminant inequality instead of an equation. Solve to find the range of m values that makes the condition true.

Exam Tip: Remember that inequalities reverse direction only when multiplying or dividing by a negative number - here we divide by positive 24, so the inequality direction stays the same.

Question 13. Find the value(s) of k for which each of the following quadratic equation has equal roots. Also, find the roots for that value(s) of k in each case.
(i) \( 3kx^2 = 4(kx - 1) \)
(ii) \( (k + 4)x^2 + (k + 1)x + 1 = 0 \)
Answer:
(i) Rearrange: \( 3kx^2 - 4kx + 4 = 0 \). With \( a = 3k \), \( b = -4k \), \( c = 4 \), for equal roots:
\( (-4k)^2 - 4(3k)(4) = 0 \)
\( 16k^2 - 48k = 0 \)
\( 16k(k - 3) = 0 \)
So \( k = 0 \) or \( k = 3 \). Since \( k = 0 \) would make the equation linear (not quadratic), we take \( k = 3 \). The equation becomes \( 9x^2 - 12x + 4 = 0 \), or \( (3x - 2)^2 = 0 \), giving the equal root \( x = \frac{2}{3} \).

(ii) For equal roots with \( a = k + 4 \), \( b = k + 1 \), \( c = 1 \):
\( (k + 1)^2 - 4(k + 4)(1) = 0 \)
\( k^2 + 2k + 1 - 4k - 16 = 0 \)
\( k^2 - 2k - 15 = 0 \)
\( (k - 5)(k + 3) = 0 \)
So \( k = 5 \) or \( k = -3 \). For \( k = 5 \): the equation is \( 9x^2 + 6x + 1 = 0 \), or \( (3x + 1)^2 = 0 \), giving \( x = -\frac{1}{3} \). For \( k = -3 \): the equation is \( x^2 - 2x + 1 = 0 \), or \( (x - 1)^2 = 0 \), giving \( x = 1 \).
In simple words: Set the discriminant to zero to find k, then substitute that value back into the original equation and solve for the root.

Exam Tip: When k makes the leading coefficient zero, the equation is no longer quadratic - exclude such values unless the problem allows it.

Question 14. Find two natural numbers which differ by 3 and whose squares have the sum 117.
Answer: Assign the first number as x. Because the difference between the two numbers is 3, the second number becomes (x + 3). We are told the sum of their squares equals 117.

\[ x^2 + (x + 3)^2 = 117 \]

\[ \implies x^2 + x^2 + 9 + 6x = 117 \]

\[ \implies 2x^2 + 6x + 9 - 117 = 0 \]

\[ \implies 2x^2 + 6x - 108 = 0 \]

\[ \implies 2(x^2 + 3x - 54) = 0 \]

\[ \implies x^2 + 3x - 54 = 0 \]

\[ \implies x^2 + 9x - 6x - 54 = 0 \]

\[ \implies x(x + 9) - 6(x + 9) = 0 \]

\[ \implies (x - 6)(x + 9) = 0 \]

\[ \implies x = 6 \text{ or } x = -9 \]

Since the numbers are natural, x must be positive, so x ≠ -9. Therefore, x = 6, and x + 3 = 9. The two required numbers are 6 and 9.
In simple words: Two numbers that are 3 apart have squares adding to 117. Solving the equation gives us 6 and 9.

Exam Tip: Always check that both solutions satisfy the original problem conditions - here, natural numbers means rejecting negative values. Verify: 6² + 9² = 36 + 81 = 117 ✓

Question 15. Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
Answer: Let the larger number be x, making the smaller number 16 - x. We need twice the square of the larger part to exceed the square of the smaller part by 164.

\[ 2x^2 - (16 - x)^2 = 164 \]

\[ \implies 2x^2 - (256 + x^2 - 32x) = 164 \]

\[ \implies 2x^2 - 256 - x^2 + 32x - 164 = 0 \]

\[ \implies x^2 + 32x - 420 = 0 \]

\[ \implies x^2 + 42x - 10x - 420 = 0 \]

\[ \implies x(x + 42) - 10(x + 42) = 0 \]

\[ \implies (x - 10)(x + 42) = 0 \]

\[ \implies x = 10 \text{ or } x = -42 \]

Since the numbers must be natural, x ≠ -42. Therefore, x = 10 and 16 - x = 6. The two required numbers are 10 and 6.
In simple words: We split 16 into two parts where twice the bigger part's square is 164 more than the smaller part's square. The answer is 10 and 6.

Exam Tip: Always verify by substitution: 2(10)² - (6)² = 2(100) - 36 = 200 - 36 = 164 ✓

Question 16. Two natural numbers are in the ratio 3 : 4. Find the numbers if the difference between their squares is 175.
Answer: Because the numbers are in the ratio 3 : 4, we can express them as 3x and 4x. The difference between their squares is 175.

\[ (4x)^2 - (3x)^2 = 175 \]

\[ \implies 16x^2 - 9x^2 = 175 \]

\[ \implies 7x^2 = 175 \]

\[ \implies x^2 = 25 \]

\[ \implies x^2 - 25 = 0 \]

\[ \implies (x - 5)(x + 5) = 0 \]

\[ \implies x = 5 \text{ or } x = -5 \]

Since the numbers are natural, x ≠ -5. Therefore, x = 5, which gives 3x = 15 and 4x = 20. The two required numbers are 15 and 20.
In simple words: Two numbers in the ratio 3 : 4 have a difference of squares equal to 175. Solving gives us 15 and 20.

Exam Tip: Remember to set up the ratio correctly - if numbers are in ratio a : b, express them as ax and bx. Verify: 20² - 15² = 400 - 225 = 175 ✓

Question 17. Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 sq. cm. Express this as an algebraic equation and solve it to find the sides of the squares.
Answer: The area of a square equals the side squared. So the first square has area x² and the second has area (x + 4)². We know their combined area is 656 cm².

\[ x^2 + (x + 4)^2 = 656 \]

\[ \implies x^2 + x^2 + 16 + 8x = 656 \]

\[ \implies 2x^2 + 8x + 16 - 656 = 0 \]

\[ \implies 2x^2 + 8x - 640 = 0 \]

\[ \implies 2(x^2 + 4x - 320) = 0 \]

\[ \implies x^2 + 4x - 320 = 0 \]

\[ \implies x^2 + 20x - 16x - 320 = 0 \]

\[ \implies x(x + 20) - 16(x + 20) = 0 \]

\[ \implies (x + 20)(x - 16) = 0 \]

\[ \implies x = -20 \text{ or } x = 16 \]

Since length cannot be negative, x ≠ -20. Therefore, x = 16 and x + 4 = 20. The sides of the two squares are 16 cm and 20 cm.
In simple words: Two squares with sides differing by 4 cm have a total area of 656 cm². Their sides are 16 cm and 20 cm.

Exam Tip: Always express the problem as an equation first, then expand and simplify before factoring. Verify: 16² + 20² = 256 + 400 = 656 ✓

Question 18. The length of a rectangular garden is 12m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.
Answer: Let the breadth be x metres, so the length is (x + 12) metres. The perimeter equals 2(length + breadth) = 2(x + x + 12) = 2(2x + 12) = (4x + 24) metres. The area equals length × breadth = x(x + 12) = (x² + 12x) m². We are told the area equals 4 times the perimeter.

\[ x^2 + 12x = 4(4x + 24) \]

\[ \implies x^2 + 12x = 16x + 96 \]

\[ \implies x^2 + 12x - 16x - 96 = 0 \]

\[ \implies x^2 - 4x - 96 = 0 \]

\[ \implies x^2 - 12x + 8x - 96 = 0 \]

\[ \implies x(x - 12) + 8(x - 12) = 0 \]

\[ \implies (x - 12)(x + 8) = 0 \]

\[ \implies x = 12 \text{ or } x = -8 \]

Since breadth cannot be negative, x ≠ -8. Therefore, x = 12 and x + 12 = 24. The length of the garden is 24 m and the breadth is 12 m.
In simple words: A rectangular garden's area equals 4 times its perimeter. If length is 12 m more than breadth, then the length is 24 m and breadth is 12 m.

Exam Tip: Check the condition carefully - here area equals 4 times the perimeter value, not 4 times each dimension. Verify: Area = 24 × 12 = 288; Perimeter = 2(24 + 12) = 72; 288 = 4 × 72 ✓

Question 19. A farmer wishes to grow a 100 m² rectangular vegetable garden. Since he has with him only 30m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.
Answer: Let x metres be the length of the side opposite the unfenced side (the wall). The length of each of the two other sides is \( \frac{1}{2}(30 - x) \). According to the problem:

\[ x \times \frac{1}{2}(30 - x) = 100 \]

\[ \implies 15x - \frac{x^2}{2} = 100 \]

\[ \implies \frac{30x - x^2}{2} = 100 \]

\[ \implies 30x - x^2 = 200 \]

\[ \implies x^2 - 30x + 200 = 0 \]

\[ \implies x^2 - 20x - 10x + 200 = 0 \]

\[ \implies x(x - 20) - 10(x - 20) = 0 \]

\[ \implies (x - 10)(x - 20) = 0 \]

\[ \implies x = 10 \text{ or } x = 20 \]

If x = 10: \( \frac{1}{2}(30 - 10) = 10 \)

If x = 20: \( \frac{1}{2}(30 - 20) = 5 \)

The dimensions of the garden are either 10 m × 10 m or 20 m × 5 m.
In simple words: The farmer has 30 m of wire to fence three sides. The area must be 100 m². The garden can be either 10 m by 10 m or 20 m by 5 m.

Exam Tip: Remember that only three sides are fenced, so the perimeter equation uses only three sides. Both solutions are valid here - verify: 10 × 10 = 100 ✓ and 20 × 5 = 100 ✓

Question 20. The hypotenuse of a right angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.
Answer: Let the shortest side be x metres. According to the problem, the third side equals (x + 1) m and the hypotenuse equals (2x - 1) m. Using the Pythagorean theorem for right angled triangles:

\[ (2x - 1)^2 = (x + 1)^2 + x^2 \]

\[ \implies 4x^2 + 1 - 4x = x^2 + 1 + 2x + x^2 \]

\[ \implies 4x^2 + 1 - 4x = 2x^2 + 2x + 1 \]

\[ \implies 4x^2 - 2x^2 - 4x - 2x + 1 - 1 = 0 \]

\[ \implies 2x^2 - 6x = 0 \]

\[ \implies 2x(x - 3) = 0 \]

\[ \implies x = 0 \text{ or } x = 3 \]

Since a side's length cannot equal zero, x ≠ 0. Therefore, x = 3, which gives (x + 1) = 4 and (2x - 1) = 5. The sides of the triangle are: shortest side = 3 m, third side = 4 m, hypotenuse = 5 m.
In simple words: In a right triangle, if the shortest side is 3 m, the next side is 4 m, and the hypotenuse is 5 m, all the given conditions are satisfied.

Exam Tip: Verify using Pythagorean theorem: 3² + 4² = 9 + 16 = 25 = 5² ✓ Also check: hypotenuse 5 is 1 less than twice 3, which is 6 ✓

Question 21. A wire, 112 cm long, is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.
Answer: The perimeter of the triangle equals the wire's total length. Sum of length of the other two sides plus hypotenuse = 112 cm. Therefore, sum of length of the other two sides = 112 - 50 = 62 cm. Let the perpendicular's length be x cm, making the base's length = (62 - x) cm. Using the Pythagorean theorem:

\[ x^2 + (62 - x)^2 = (50)^2 \]

\[ \implies x^2 + 3844 + x^2 - 124x = 2500 \]

\[ \implies 2x^2 - 124x + 3844 - 2500 = 0 \]

\[ \implies 2x^2 - 124x + 1344 = 0 \]

\[ \implies 2(x^2 - 62x + 672) = 0 \]

\[ \implies x^2 - 62x + 672 = 0 \]

\[ \implies x^2 - 48x - 14x + 672 = 0 \]

\[ \implies x(x - 48) - 14(x - 48) = 0 \]

\[ \implies (x - 48)(x - 14) = 0 \]

\[ \implies x = 48 \text{ or } x = 14 \]

The two sides are 48 cm and 14 cm. The area of the triangle:

\[ \text{Area} = \frac{1}{2} \times \text{Perpendicular} \times \text{Base} = \frac{1}{2} \times 48 \times 14 = 336 \text{ cm}^2 \]
In simple words: A 112 cm wire forms a right triangle with hypotenuse 50 cm. The other two sides are 48 cm and 14 cm, giving an area of 336 cm².

Exam Tip: Always verify Pythagorean theorem: 48² + 14² = 2304 + 196 = 2500 = 50² ✓ Remember to use both roots to find the two perpendicular sides.

Question 22. The speed of a boat in still water is 11 km/h. It can go 12 km upstream and return downstream to original point in 2 hours 45 minutes. Find the speed of the stream.
Answer: Let the speed of the stream be x km/h. The boat's speed in still water is 11 km/h, so its upstream speed = (11 - x) km/h and downstream speed = (11 + x) km/h. The total time for 12 km upstream plus 12 km downstream is 2 hours 45 minutes.

Converting 2 hours 45 minutes to hours: 2 hours 45 minutes = \( \frac{165}{60} \) hours = \( \frac{11}{4} \) hours

\[ \frac{12}{11 - x} + \frac{12}{11 + x} = \frac{11}{4} \]

\[ \implies \frac{12(11 + x) + 12(11 - x)}{(11 + x)(11 - x)} = \frac{11}{4} \]

\[ \implies \frac{132 + 12x + 132 - 12x}{121 - x^2} = \frac{11}{4} \]

\[ \implies \frac{264}{121 - x^2} = \frac{11}{4} \]

\[ \implies 264 \times 4 = 11(121 - x^2) \]

\[ \implies 1056 = 1331 - 11x^2 \]

\[ \implies 11x^2 = 1331 - 1056 \]

\[ \implies 11x^2 = 275 \]

\[ \implies x^2 = 25 \]

\[ \implies x^2 - 25 = 0 \]

\[ \implies (x - 5)(x + 5) = 0 \]

\[ \implies x = 5 \text{ or } x = -5 \]

Since speed cannot be negative, x ≠ -5. Therefore, the speed of the stream is 5 km/h.
In simple words: A boat travels 12 km upstream and 12 km downstream in 2 hours 45 minutes. With still-water speed of 11 km/h, the stream's speed is 5 km/h.

Exam Tip: Always convert time units carefully - here 2 hours 45 minutes = 165 minutes = 11/4 hours. Verify: Time upstream = 12/(11-5) = 2 hours; Time downstream = 12/(11+5) = 0.75 hours; Total = 2.75 = 2 hours 45 minutes ✓

Question 23. A man spent Rs2800 on buying a number of plants priced at Rs x each. Because of the number involved, the supplier reduced the price of each plant by one rupee. The man finally paid Rs2730 and received 10 more plants. Find x.
Answer: Let the initial price per plant be Rs x. Then the number of plants bought for Rs2800 is \( \frac{2800}{x} \).

After the price reduction of Rs 1 per plant, the new price becomes Rs (x - 1). The man now buys \( \frac{2800}{x} + 10 \) plants for Rs2730.

Setting up the equation based on the given condition:

\( \left(\frac{2800}{x} + 10\right)(x - 1) = 2730 \)

\( \Rightarrow \frac{(2800 + 10x)(x - 1)}{x} = 2730 \)

\( \Rightarrow (2800 + 10x)(x - 1) = 2730x \)

\( \Rightarrow 2800x - 2800 + 10x^2 - 10x = 2730x \)

\( \Rightarrow 10x^2 + 2790x - 2800 = 2730x \)

\( \Rightarrow 10x^2 + 60x - 2800 = 0 \)

\( \Rightarrow x^2 + 6x - 280 = 0 \)

\( \Rightarrow x^2 + 20x - 14x - 280 = 0 \)

\( \Rightarrow x(x + 20) - 14(x + 20) = 0 \)

\( \Rightarrow (x - 14)(x + 20) = 0 \)

Thus, \( x = 14 \) or \( x = -20 \).

Since price cannot be negative, \( x = 14 \).

In simple words: The price of each plant was Rs 14. With the discount, he bought more plants for less total money.

Exam Tip: Always reject negative solutions when the variable represents a physical quantity like price or distance - only the positive, practical answer is valid.

Question 24. Forty years hence, Mr. Pratap's age will be the square of what it was 32 years ago. Find his present age.
Answer: Let Mr. Pratap's current age be x years.

After 40 years, his age will be (x + 40) years.

32 years ago, his age was (x - 32) years.

According to the given condition:

\( (x + 40) = (x - 32)^2 \)

\( \Rightarrow x + 40 = x^2 - 64x + 1024 \)

\( \Rightarrow x^2 - 64x - x + 1024 - 40 = 0 \)

\( \Rightarrow x^2 - 65x + 984 = 0 \)

\( \Rightarrow x^2 - 24x - 41x + 984 = 0 \)

\( \Rightarrow x(x - 24) - 41(x - 24) = 0 \)

\( \Rightarrow (x - 41)(x - 24) = 0 \)

Thus, \( x = 41 \) or \( x = 24 \).

Since x must be at least 32 for the age "32 years ago" to be positive, \( x \neq 24 \).

Therefore, \( x = 41 \).

Mr. Pratap's present age is 41 years.

In simple words: In 40 years, Mr. Pratap will be 81, which is \( 9^2 \). Nine years ago (32 years back from now), he was 9.

Exam Tip: Always check whether both roots satisfy the practical conditions of the problem - one solution may be mathematically valid but contextually impossible.

Question 25. The total expenses of a trip for certain number of people is Rs 18,000. If three more people join them, then the share of each reduces by Rs 3,000. Take x to be the original number of people, form a quadratic equation in x and solve it to find the value of x.
Answer: Let the original number of people be x.

Total trip expense = Rs 18,000.

Expense per person initially = \( \frac{18000}{x} \)

When 3 more people join, the number becomes (x + 3).

New expense per person = \( \frac{18000}{x + 3} \)

According to the given condition, the share reduces by Rs 3,000:

\( \frac{18000}{x} - \frac{18000}{x + 3} = 3000 \)

\( \Rightarrow \frac{18000(x + 3) - 18000x}{x(x + 3)} = 3000 \)

\( \Rightarrow \frac{18000x + 54000 - 18000x}{x(x + 3)} = 3000 \)

\( \Rightarrow \frac{54000}{x(x + 3)} = 3000 \)

\( \Rightarrow \frac{54000}{3000} = x(x + 3) \)

\( \Rightarrow 18 = x^2 + 3x \)

\( \Rightarrow x^2 + 3x - 18 = 0 \)

\( \Rightarrow x^2 + 6x - 3x - 18 = 0 \)

\( \Rightarrow x(x + 6) - 3(x + 6) = 0 \)

\( \Rightarrow (x - 3)(x + 6) = 0 \)

Thus, \( x = 3 \) or \( x = -6 \).

Since the number of people cannot be negative, \( x = 3 \).

The original number of people is 3.

In simple words: There were 3 people sharing Rs 18,000. Each paid Rs 6,000. When 3 more joined (6 total), each paid Rs 3,000 - exactly Rs 3,000 less.

Exam Tip: Always verify your answer by substituting back into the original condition - it takes only 10 seconds and confirms your solution is correct.

Question 26. A car travels a distance of 72 km at a certain average speed of x km per hour and then travels a distance of 81 km at an average speed of 6 km per hour more than its original average speed. If it takes 3 hours to complete the total journey then form a quadratic equation and solve it to find its original average speed.
Answer: Let the original average speed be x km per hour.

The car travels 72 km at x km/hr, taking time = \( \frac{72}{x} \) hours.

It then travels 81 km at (x + 6) km/hr, taking time = \( \frac{81}{x + 6} \) hours.

Total time for the journey = 3 hours.

Therefore:

\( \frac{72}{x} + \frac{81}{x + 6} = 3 \)

\( \Rightarrow \frac{72(x + 6) + 81x}{x(x + 6)} = 3 \)

\( \Rightarrow 72(x + 6) + 81x = 3x(x + 6) \)

\( \Rightarrow 72x + 432 + 81x = 3x^2 + 18x \)

\( \Rightarrow 153x + 432 = 3x^2 + 18x \)

\( \Rightarrow 3x^2 + 18x - 153x - 432 = 0 \)

\( \Rightarrow 3x^2 - 135x - 432 = 0 \)

\( \Rightarrow 3(x^2 - 45x - 144) = 0 \)

\( \Rightarrow x^2 - 45x - 144 = 0 \)

\( \Rightarrow x^2 - 48x + 3x - 144 = 0 \)

\( \Rightarrow x(x - 48) + 3(x - 48) = 0 \)

\( \Rightarrow (x + 3)(x - 48) = 0 \)

Thus, \( x = -3 \) or \( x = 48 \).

Since speed must be positive, \( x = 48 \).

The original average speed is 48 km/hr.

In simple words: The car's first part of the trip was slow at 48 km/hr. The second part was faster at 54 km/hr. Together, the two parts took exactly 3 hours.

Exam Tip: When forming time equations, always use time = distance / speed, then sum all time segments to match the total time given.