CBSE Class 10 Mathematics Quadratic Equations Assignment Set 09

Question 1. The general form of a quadratic equation is (\(a \neq 0\))
(a) \(ax^2 + bx + c\)
(b) \(ax^2 + bx + c = 0\)
(c) \(ax + b\)
(d) \(ax + b = 0\)
Answer: (b) \(ax^2 + bx + c = 0\)
A quadratic equation is a second-order polynomial equation in a single variable \(x\) with \(a \neq 0\), set equal to zero.
In simple words: The standard equation has an equal sign and a squared term.

Exam Tip: Always verify that \(a \neq 0\); otherwise, it becomes a linear equation.

Question 2. Number of solutions of a quadratic equation are :
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (c) 2
According to the Fundamental Theorem of Algebra, a polynomial equation of degree 2 has exactly two roots (real or complex).
In simple words: Quadratic equations always have two roots or solutions.

Exam Tip: The two roots can be distinct real, equal real, or non-real complex numbers.

Question 3. If the equation \(x^2 - (2 + m) x + (-m^2 - 4m - 4) = 0\) has coincident roots, then
(a) \(m = 0, m = 1\)
(b) \(m = 2, m = 2\)
(c) \(m = -2, m = -2\)
(d) \(m = 6, m = 1\)
Answer: (c) \(m = -2, m = -2\)
For coincident roots, the discriminant \(D = 0\).
Here, \(a = 1\), \(b = -(2 + m)\), \(c = -(m^2 + 4m + 4) = -(m+2)^2\).
\(D = b^2 - 4ac = [-(2+m)]^2 - 4(1)[-(m+2)^2] = 0\)
\((m+2)^2 + 4(m+2)^2 = 0 \implies 5(m+2)^2 = 0 \implies m = -2\).
In simple words: Coincident roots mean discriminant is zero, giving \(m = -2\).

Exam Tip: Always factorize quadratic expressions inside the discriminant to simplify calculations.

Question 4. Discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is given by
(a) \(\sqrt{b^2 - 4ac}\)
(b) \(\sqrt{b^2 + 4ac}\)
(c) \(b^2 - 4ac\)
(d) \(b^2 + 4ac\)
Answer: (c) \(b^2 - 4ac\)
The discriminant \(D\) is defined as \(b^2 - 4ac\) and determines the nature of the roots.
In simple words: The formula for the discriminant is \(b^2 - 4ac\).

Exam Tip: Do not confuse the discriminant formula with the quadratic formula containing the square root.

Question 5. Which is a quadratic equation?
(a) \(x + \frac{1}{x} = 2\)
(b) \(x^2 + 1 = (x + 3)^2\)
(c) \(x(x + 2) = x(x + 2)\)
(d) \(x + \frac{1}{x} = x^2\)
Answer: (a) \(x + \frac{1}{x} = 2\)
Multiplying \(x + \frac{1}{x} = 2\) by \(x\) gives \(x^2 + 1 = 2x\) or \(x^2 - 2x + 1 = 0\), which is quadratic.
Option (b) simplifies to \(x^2 + 1 = x^2 + 6x + 9 \implies 6x + 8 = 0\) (linear).
In simple words: Simplifying option (a) results in a power of 2 for \(x\).

Exam Tip: Always simplify rational equations by clearing denominators before checking the degree.

Question 6. If the roots of a quadratic equation are 2 and 3, then the equation is
(a) \(x^2 + 5x + 6 = 0\)
(b) \(x^2 + 5x - 6 = 0\)
(c) \(x^2 - 5x - 6 = 0\)
(d) \(x^2 - 5x + 6 = 0\)
Answer: (d) \(x^2 - 5x + 6 = 0\)
Using the formula \(x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0\).
Sum = \(2 + 3 = 5\), Product = \(2 \times 3 = 6\).
Equation: \(x^2 - 5x + 6 = 0\).
In simple words: Subtract the sum of roots from \(x\) coefficient and add the product as the constant term.

Exam Tip: Remember the structure: \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\).

Question 7. Roots of the equations \(x^2 - 3x + 2 = 0\) are
(a) \(1, -2\)
(b) \(-1, 2\)
(c) \(-1, -2\)
(d) \(1, 2\)
Answer: (d) \(1, 2\)
Factorizing \(x^2 - 3x + 2 = 0\):
\(x^2 - 2x - x + 2 = 0\)
\(x(x - 2) - 1(x - 2) = 0 \implies (x - 1)(x - 2) = 0\)
Roots are \(x = 1\) and \(x = 2\).
In simple words: Splitting the middle term gives factors \((x-1)\) and \((x-2)\).

Exam Tip: Verify roots by substituting them back into the original equation.

Question 8. If the roots of a quadratic equation are equal, than discriminant is
(a) 1
(b) 0
(c) greater than 0
(d) less than zero.
Answer: (b) 0
Equal roots occur if and only if the discriminant \(b^2 - 4ac = 0\).
In simple words: Equal roots mean the discriminant is exactly zero.

Question 9. If one root of \(2x^2 + kx + 1 = 0\) is \(-\frac{1}{2}\), then the value of 'k' is
(a) 3
(b) \(-3\)
(c) 5
(d) \(-5\)
Answer: (a) 3
Substitute \(x = -\frac{1}{2}\) into the equation:
\(2\left(-\frac{1}{2}\right)^2 + k\left(-\frac{1}{2}\right) + 1 = 0\)
\(2\left(\frac{1}{4}\right) - \frac{k}{2} + 1 = 0 \implies \frac{1}{2} - \frac{k}{2} + 1 = 0\)
\(\frac{3}{2} = \frac{k}{2} \implies k = 3\).
In simple words: Plug the given root into the equation and solve for \(k\).

Exam Tip: A root always satisfies its equation when substituted for the variable.

Question 10. The sum of the roots of the quadratic \(5x^2 - 6x + 1 = 0\) is
(a) \(\frac{6}{5}\)
(b) \(\frac{1}{5}\)
(c) \(-\frac{5}{6}\)
(d) \(-\frac{1}{5}\)
Answer: (a) \(\frac{6}{5}\)
Sum of roots = \(-\frac{b}{a}\). Here \(a = 5\), \(b = -6\), \(c = 1\).
Sum = \(-\frac{-6}{5} = \frac{6}{5}\).
In simple words: Sum of roots is negative of coefficient of \(x\) divided by coefficient of \(x^2\).

Exam Tip: Remember the relation: Sum = \(-b/a\) and Product = \(c/a\).

Question 11. The product of the roots of the quadratic equation \(2x^2 + 5x - 7 = 0\) is
(a) \(\frac{5}{2}\)
(b) \(-\frac{7}{2}\)
(c) \(-\frac{5}{2}\)
(d) \(\frac{7}{2}\)
Answer: (b) \(-\frac{7}{2}\)
Product of roots = \(\frac{c}{a} = \frac{-7}{2} = -\frac{7}{2}\).
In simple words: Product of roots is constant term divided by coefficient of \(x^2\).

Exam Tip: Always include the correct sign of the constant term and leading coefficient.

Question 12. If the roots of the quadratic \(2x^2 + kx + 2 = 0\) are equal then the value of 'k' is
(a) 4
(b) \(-4\)
(c) \(\pm 4\)
(d) \(\pm 16\)
Answer: (c) \(\pm 4\)
For equal roots, \(D = 0 \implies b^2 - 4ac = 0\).
\(k^2 - 4(2)(2) = 0 \implies k^2 - 16 = 0 \implies k^2 = 16 \implies k = \pm 4\).
In simple words: Setting discriminant to zero gives \(k^2 = 16\), so \(k\) can be positive or negative 4.

Exam Tip: Do not forget both positive and negative square roots when solving \(k^2 = c\).

Question 13. If the roots of \(4x^2 + 3px + 9 = 0\) are real and distinct then, the value of \(p\) is
(a) \(p \geq -4 \text{ or } p \leq 4\)
(b) \(p < -4 \text{ or } p > 4\)
(c) \(p \leq -4 \text{ or } p \leq 4\)
(d) \(p \leq -4 \text{ or } p \geq 4\)
Answer: (b) \(p < -4 \text{ or } p > 4\)
For real and distinct roots, \(D > 0\).
\((3p)^2 - 4(4)(9) > 0 \implies 9p^2 - 144 > 0 \implies 9(p^2 - 16) > 0\)
\(p^2 > 16 \implies p < -4 \text{ or } p > 4\).
In simple words: Real and distinct roots require the discriminant to be strictly greater than zero.

Exam Tip: Inequality \(p^2 > a^2\) splits into \(p < -a\) or \(p > a\).

Question 14. If the sum and product of roots of a quadratic equation are \(-\frac{7}{2}\) and \(\frac{5}{2}\) respectively, then the equation is
(a) \(2x^2 + 7x + 5 = 0\)
(b) \(2x^2 - 7x + 5 = 0\)
(c) \(2x^2 - 7x - 5 = 0\)
(d) \(2x^2 + 7x - 5 = 0\)
Answer: (a) \(2x^2 + 7x + 5 = 0\)
Equation format: \(x^2 - (\text{Sum})x + \text{Product} = 0\)
\(x^2 - \left(-\frac{7}{2}\right)x + \frac{5}{2} = 0 \implies x^2 + \frac{7}{2}x + \frac{5}{2} = 0\)
Multiplying by 2 gives \(2x^2 + 7x + 5 = 0\).
In simple words: Substitute sum and product into standard form and clear the denominator.

Exam Tip: Multiply through by the denominator to match standard integer coefficient forms.

Question 15. Which constant must be added or subtracted to solve the equation \(9x^2 + \frac{3}{4}x - \sqrt{2} = 0\) by the method of completing the square
(a) \(\frac{1}{8}\)
(b) \(\frac{1}{64}\)
(c) \(\frac{1}{16}\)
(d) none
Answer: (b) \(\frac{1}{64}\)
Divide by 9: \(x^2 + \frac{3}{36}x - \frac{\sqrt{2}}{9} = 0 \implies x^2 + \frac{1}{12}x - \frac{\sqrt{2}}{9} = 0\).
Coefficient of \(x\) is \(\frac{1}{12}\). Take half of it: \(\frac{1}{2} \times \frac{1}{12} = \frac{1}{24}\).
Square it: \(\left(\frac{1}{24}\right)^2 = \frac{1}{576}\)... wait, re-evaluating coefficient: equation has \(\frac{3}{4}x\). Divided by 9 gives \(\frac{3}{36}x = \frac{1}{12}x\). Half is \(\frac{1}{24}\), square is \(\frac{1}{576}\). Let's check option values: option (b) is \(\frac{1}{64}\). If the term is \(\frac{3}{4}x\) and coefficient of \(x^2\) is 9, rewriting as \((3x)^2 + 2(3x)\left(\frac{1}{8}\right) + \left(\frac{1}{8}\right)^2 - \left(\frac{1}{8}\right)^2 - \sqrt{2} = 0\), the term to add/subtract is \(\left(\frac{1}{8}\right)^2 = \frac{1}{64}\).
In simple words: Express \(9x^2\) as \((3x)^2\) and find the second term to complete the square.

Exam Tip: Square half of the coefficient of the middle term after making the leading coefficient a square.

SHORT ANSWER TYPE QUESTIONS

Question 16. If one root of the equation \(x^2 + 7x + k = 0\) is \(-2\), then find the value of \(k\) and the other root.
Answer:
Substitute \(x = -2\) into the equation:
\((-2)^2 + 7(-2) + k = 0\)
\(4 - 14 + k = 0 \implies -10 + k = 0 \implies k = 10\)
Substitute \(k = 10\) back into the equation: \(x^2 + 7x + 10 = 0\)
Factorize: \((x + 5)(x + 2) = 0\)
The roots are \(x = -2\) and \(x = -5\).
Thus, \(k = 10\) and the other root is \(-5\).
In simple words: Substitute the known root to find \(k\), then factorize to find the second root.

Exam Tip: Always state both the value of \(k\) and the other root clearly.

Question 17. For what value of 'k' the equation \(2x^2 + kx + 3 = 0\) has equal roots?
Answer:
For equal roots, discriminant \(D = 0\).
\(b^2 - 4ac = 0\)
\(k^2 - 4(2)(3) = 0\)
\(k^2 - 24 = 0 \implies k^2 = 24\)
\(k = \pm \sqrt{24} = \pm 2\sqrt{6}\).
In simple words: Set \(b^2 - 4ac = 0\) and solve for \(k\).

Exam Tip: Simplify square roots completely (\(\sqrt{24} = 2\sqrt{6}\)).

Question 18. For what value of 'p', the equation \(3x^2 + px + 3 = 0\) has real roots?
Answer:
For real roots, \(D \geq 0\).
\(p^2 - 4(3)(3) \geq 0\)
\(p^2 - 36 \geq 0 \implies p^2 \geq 36\)
\(p \geq 6 \text{ or } p \leq -6\).
In simple words: Real roots require the discriminant to be greater than or equal to zero.

Exam Tip: Inequality \(p^2 \geq a^2\) means \(p \geq a\) or \(p \leq -a\).

Question 19. The product of two consecutive odd integers is 63. Represent this in form of a quadratic equation.
Answer:
Let the first odd integer be \(x\).
The next consecutive odd integer is \(x + 2\).
According to the question, their product is 63:
\(x(x + 2) = 63\)
\(x^2 + 2x - 63 = 0\).
In simple words: Assume integers as \(x\) and \(x+2\) and multiply them to get 63.

Exam Tip: Consecutive odd (or even) integers always differ by 2.

Question 20. Find the roots of the equation : \(x + \frac{1}{x} = 4\frac{1}{4}\), \(x \neq 0\).
Answer:
Rewrite the equation:
\(x + \frac{1}{x} = \frac{17}{4}\)
Multiply through by \(4x\):
\(4x^2 + 4 = 17x \implies 4x^2 - 17x + 4 = 0\)
Factorize by splitting middle term:\(-16x - x\):
\(4x(x - 4) - 1(x - 4) = 0\)
\((4x - 1)(x - 4) = 0\)
Roots are \(x = \frac{1}{4}\) and \(x = 4\).
In simple words: Clear denominators to form a standard quadratic equation and factorize.

Exam Tip: Mixed fractions like \(4\frac{1}{4}\) must be converted to improper fractions first.

Question 21. Find the roots of the equation : \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\).
Answer:
Product of \(a\) and \(c\) = \(\sqrt{2} \times 5\sqrt{2} = 10\).
Split the middle term \(7x\) into \(5x + 2x\):
\(\sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0\)
\(\sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0\)
\((\sqrt{2}x + 5)(x + \sqrt{2}) = 0\)
Roots are \(x = -\frac{5}{\sqrt{2}}\) and \(x = -\sqrt{2}\).
In simple words: Split middle term \(7x\) into \(5x\) and \(2x\) using the product of outer coefficients.

Exam Tip: Recognize that \(2 = \sqrt{2} \times \sqrt{2}\) to simplify factorization with square roots.

Question 22. Divide 51 in to two parts such that their product is 378.
Answer:
Let one part be \(x\), then the other part is \(51 - x\).
Product = \(x(51 - x) = 378\)
\(51x - x^2 = 378 \implies x^2 - 51x + 378 = 0\)
Factorize by splitting middle term \(-42x - 9x\):
\(x(x - 42) - 9(x - 42) = 0\)
\((x - 9)(x - 42) = 0\)
The parts are 9 and 42.
In simple words: Let parts be \(x\) and \(51-x\), form the quadratic equation, and solve.

Exam Tip: Factoring large numbers can be done by finding prime factors of 378.

Question 23. Find 'k' so that \((k - 12) x^2 + 2 (k - 12) x + 2 = 0\) has equal roots. (\(k \neq 12\)).
Answer:
For equal roots, \(D = 0 \implies b^2 - 4ac = 0\).
Here \(a = k - 12\), \(b = 2(k - 12)\), \(c = 2\).
\([2(k - 12)]^2 - 4(k - 12)(2) = 0\)
\(4(k - 12)^2 - 8(k - 12) = 0\)
Factor out \(4(k - 12)\):
\(4(k - 12)[(k - 12) - 2] = 0\)
\(4(k - 12)(k - 14) = 0\)
Since \(k \neq 12\), we have \(k - 14 = 0 \implies k = 14\).
In simple words: Substitute coefficients into the discriminant formula and factor out common terms.

Exam Tip: Always check excluded values given in the question (\(k \neq 12\)).

Question 24. If \((-5)\) is a root of the equation \(2x^2 + px - 15 = 0\) and the equation \(p(x^2 + x) + k = 0\) has equal roots, find values of \(p\) and \(k\).
Answer:
Substitute \(x = -5\) into \(2x^2 + px - 15 = 0\):
\(2(-5)^2 + p(-5) - 15 = 0\)
\(2(25) - 5p - 15 = 0 \implies 50 - 5p - 15 = 0\)
\(35 - 5p = 0 \implies 5p = 35 \implies p = 7\).
Now substitute \(p = 7\) into \(p(x^2 + x) + k = 0\):
\(7(x^2 + x) + k = 0 \implies 7x^2 + 7x + k = 0\).
Since this equation has equal roots, \(D = 0\):
\(7^2 - 4(7)(k) = 0 \implies 49 - 28k = 0\)
\(28k = 49 \implies k = \frac{49}{28} = \frac{7}{4}\).
Thus, \(p = 7\) and \(k = \frac{7}{4}\).
In simple words: Find \(p\) using the given root, then use discriminant equals zero to find \(k\).

Exam Tip: Solve step-by-step by finding parameters sequentially.

Question 25. Find the roots of the equation \(\frac{1}{x + 2} - \frac{1}{x} = 3\), \(x \neq -2, x \neq 0\).
Answer:
Take the LCM on the left side:
\(\frac{x - (x + 2)}{x(x + 2)} = 3 \implies \frac{-2}{x^2 + 2x} = 3\)
\(-2 = 3(x^2 + 2x) \implies 3x^2 + 6x + 2 = 0\)
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-6 \pm \sqrt{6^2 - 4(3)(2)}}{2(3)} = \frac{-6 \pm \sqrt{36 - 24}}{6} = \frac{-6 \pm \sqrt{12}}{6}\)
\(x = \frac{-6 \pm 2\sqrt{3}}{6} = \frac{-3 \pm \sqrt{3}}{3}\).
Roots are \(\frac{-3 + \sqrt{3}}{3}\) and \(\frac{-3 - \sqrt{3}}{3}\).
In simple words: Simplify into a quadratic equation and apply the quadratic formula.

Exam Tip: Simplify radical fractions by dividing numerator and denominator by common factors.

Question 26. The difference of two numbers is 5 and the difference of their reciprocals is \(\frac{1}{10}\). Find the numbers.
Answer:
Let the smaller number be \(x\), then the larger number is \(x + 5\).
Difference of their reciprocals is \(\frac{1}{10}\):
\(\frac{1}{x} - \frac{1}{x + 5} = \frac{1}{10}\)
\(\frac{(x + 5) - x}{x(x + 5)} = \frac{1}{10} \implies \frac{5}{x^2 + 5x} = \frac{1}{10}\)
\(x^2 + 5x = 50 \implies x^2 + 5x - 50 = 0\)
Factorize: \((x + 10)(x - 5) = 0\)
Since \(x > 0\) (or taking positive numbers), \(x = 5\).
The numbers are 5 and \(5 + 5 = 10\).
In simple words: Set numbers as \(x\) and \(x+5\), set up reciprocal difference equation, and solve.

Exam Tip: Ensure the reciprocal of the smaller number comes first so the difference is positive.

Question 27. If the roots of the equation \((b - c)x^2 + (c - a) x + (a - b) = 0\) are equal, then prove that \(2b = a + c\).
Answer:
For equal roots, discriminant \(D = 0\):
\((c - a)^2 - 4(b - c)(a - b) = 0\)
\(c^2 + a^2 - 2ac - 4(ab - b^2 - ac + bc) = 0\)
\(a^2 + c^2 - 2ac - 4ab + 4b^2 + 4ac - 4bc = 0\)
\(a^2 + 4b^2 + c^2 + 2ac - 4ab - 4bc = 0\)
Notice this expands to \((a - 2b + c)^2 = 0\):
\(a - 2b + c = 0 \implies 2b = a + c\).
Hence proved.
In simple words: Expand the discriminant condition and recognize it as a perfect square.

Exam Tip: Algebraic identity expansions simplify symmetric polynomial proofs quickly.

Question 28. Find the nature of the roots of the following quadratic equations. If roots are real, find them.
(a) \(5x^2 - 3x + 2 = 0\).
(b) \(2x^2 - 9x + 9 = 0\).
Answer:
(a) For \(5x^2 - 3x + 2 = 0\):
\(D = b^2 - 4ac = (-3)^2 - 4(5)(2) = 9 - 40 = -31 < 0\).
Since \(D < 0\), the equation has no real roots.

(b) For \(2x^2 - 9x + 9 = 0\):
\(D = (-9)^2 - 4(2)(9) = 81 - 72 = 9 > 0\).
Since \(D > 0\), roots are real and distinct.
Using quadratic formula: \(x = \frac{-(-9) \pm \sqrt{9}}{2(2)} = \frac{9 \pm 3}{4}\)
\(x_1 = \frac{9 + 3}{4} = \frac{12}{4} = 3\)
\(x_2 = \frac{9 - 3}{4} = \frac{6}{4} = \frac{3}{2}\).
Roots are 3 and \(\frac{3}{2}\).
In simple words: Check the discriminant first. If negative, stop; if positive, find roots using the formula.

Exam Tip: Always calculate discriminant first to test for real roots.

Question 29. Sum of two numbers is 15, if sum of their reciprocals is \(\frac{3}{10}\). Find the numbers.
Answer:
Let one number be \(x\), then the other number is \(15 - x\).
\(\frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10}\)
\(\frac{(15 - x) + x}{x(15 - x)} = \frac{3}{10} \implies \frac{15}{15x - x^2} = \frac{3}{10}\)
Cross-multiplying: \(150 = 3(15x - x^2) \implies 50 = 15x - x^2\)
\(x^2 - 15x + 50 = 0\)
Factorize: \((x - 10)(x - 5) = 0\)
Thus, \(x = 5\) or \(x = 10\).
The numbers are 5 and 10.
In simple words: Represent numbers as \(x\) and \(15-x\), form the reciprocal equation, and factorize.

Exam Tip: Symmetry in the equation means the two numbers will be the two roots of the quadratic.

Question 30. Solve the following quadratic equations
\(2\left(\frac{2x + 3}{x - 3}\right) - 25\left(\frac{x - 3}{2x + 3}\right) = 5\), \(x \neq 3, x \neq -\frac{3}{2}\)
Answer:
Let \(y = \frac{2x + 3}{x - 3}\), then \(\frac{1}{y} = \frac{x - 3}{2x + 3}\).
Substitute into the equation:
\(2y - \frac{25}{y} = 5\)
Multiply by \(y\): \(2y^2 - 25 = 5y \implies 2y^2 - 5y - 25 = 0\)
Factorize: \(2y^2 - 10y + 5y - 25 = 0 \implies 2y(y - 5) + 5(y - 5) = 0\)
\((2y + 5)(y - 5) = 0 \implies y = -\frac{5}{2} \text{ or } y = 5\).
Case 1: If \(y = 5\), \(\frac{2x + 3}{x - 3} = 5 \implies 2x + 3 = 5x - 15 \implies 3x = 18 \implies x = 6\).
Case 2: If \(y = -\frac{5}{2}\), \(\frac{2x + 3}{x - 3} = -\frac{5}{2} \implies 4x + 6 = -5x + 15 \implies 9x = 9 \implies x = 1\).
Solutions are \(x = 6, 1\).
In simple words: Substitute substitution variable \(y\), solve for \(y\), then find \(x\).

Exam Tip: Substitution simplifies complex fractional equations into standard quadratics.

Question 31. \(a^2x^2 + (a^2 - b^2) x - b^2 = 0\).
Answer:
Using factorization by splitting the middle term \((a^2x^2 - b^2x + a^2x - b^2\) is incorrect, let's use coefficients: sum is \(a^2 - b^2\), product is \(-a^2b^2\).
The terms are \(a^2x^2\) and \(-b^2\), middle term split into \(a^2x\) and \(-b^2x\):
\(a^2x^2 + a^2x - b^2x - b^2 = 0\)
\(a^2x(x + 1) - b^2(x + 1) = 0\)
\((a^2x - b^2)(x + 1) = 0\)
Roots are \(x = -1\) and \(x = \frac{b^2}{a^2}\).
In simple words: Group terms to factorize variables with algebraic coefficients.

Exam Tip: Look for common factors after grouping adjacent pairs.

Question 32. \(4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0\).
Answer:
Product of \(a\) and \(c\) = \(4\sqrt{3} \times (-2\sqrt{3}) = -24\).
Split middle term \(5x\) into \(8x - 3x\):
\(4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0\)
\(4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0\)
\((4x - \sqrt{3})(\sqrt{3}x + 2) = 0\)
Roots are \(x = \frac{\sqrt{3}}{4}\) and \(x = -\frac{2}{\sqrt{3}}\).
In simple words: Split \(5x\) into \(8x\) and \(-3x\) to factor out common radical terms.

Exam Tip: Note that \(3 = \sqrt{3} \times \sqrt{3}\) helps in factoring terms involving \(\sqrt{3}\).

Question 33. \(ab x^2 + (b^2 - ac) x - bc = 0\).
Answer:
Expand the middle term: \(ab x^2 + b^2x - acx - bc = 0\)
Group terms: \(bx(ax + b) - c(ax + b) = 0\)
\((bx - c)(ax + b) = 0\)
Roots are \(x = \frac{c}{b}\) and \(x = -\frac{b}{a}\).
In simple words: Expand and factor by grouping terms in pairs.

Exam Tip: Grouping works well when coefficients contain multiple variables.

Question 34. \(\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = \frac{10}{3}\), \(x \neq 2, x \neq 4\).
Answer:
Take LCM on the left side:
\(\frac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \frac{10}{3}\)
\(\frac{(x^2 - 5x + 4) + (x^2 - 5x + 6)}{x^2 - 6x + 8} = \frac{10}{3}\)
\(\frac{2x^2 - 10x + 10}{x^2 - 6x + 8} = \frac{10}{3} \implies \frac{x^2 - 5x + 5}{x^2 - 6x + 8} = \frac{5}{3}\)
Cross-multiplying: \(3(x^2 - 5x + 5) = 5(x^2 - 6x + 8)\)
\(3x^2 - 15x + 15 = 5x^2 - 30x + 40\)
\(2x^2 - 15x + 25 = 0\)
Factorize: \(2x^2 - 10x - 5x + 25 = 0 \implies 2x(x - 5) - 5(x - 5) = 0\)
\((2x - 5)(x - 5) = 0\)
Roots are \(x = \frac{5}{2}\) and \(x = 5\).
In simple words: Cross-multiply after combining fractions to form a standard quadratic equation.

Exam Tip: Simplify coefficients by dividing by common factors before factorizing.

Question 35. \(\frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}\), \(x \neq -4, x \neq 7\).
Answer:
Take LCM:\(\frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30}\)
\(\frac{-11}{x^2 - 3x - 28} = \frac{11}{30}\)
Divide both sides by 11: \(\frac{-1}{x^2 - 3x - 28} = \frac{1}{30}\)
\(-30 = x^2 - 3x - 28 \implies x^2 - 3x + 2 = 0\)
Factorize: \((x - 1)(x - 2) = 0\)
Roots are \(x = 1\) and \(x = 2\).
In simple words: Subtract fractions, cancel common numerator terms, and solve the resulting quadratic.

Exam Tip: Canceling common factors on both sides simplifies numbers significantly.

Question 36. \(3x^2 + 2\sqrt{5}x - 5 = 0\).
Answer:
Product of \(a\) and \(c\) = \(3 \times (-5) = -15\).
Split middle term \(2\sqrt{5}x\) into \(3\sqrt{5}x - \sqrt{5}x\):
\(3x^2 + 3\sqrt{5}x - \sqrt{5}x - 5 = 0\)
\(3x(x + \sqrt{5}) - \sqrt{5}(x + \sqrt{5}) = 0\)
\((3x - \sqrt{5})(x + \sqrt{5}) = 0\)
Roots are \(x = \frac{\sqrt{5}}{3}\) and \(x = -\sqrt{5}\).
In simple words: Split the middle term using \(\sqrt{5}\) coefficients.

Exam Tip: Remember that \(5 = \sqrt{5} \times \sqrt{5}\) when factoring surd expressions.

Question 37. \(\frac{1}{a} + \frac{1}{b} + \frac{1}{x} = \frac{1}{a + b + x}\), \(a \neq 0, b \neq 0, x \neq 0, x \neq -(a + b)\).
Answer:
Rearrange terms: \(\frac{1}{x} - \frac{1}{a + b + x} = -\left(\frac{1}{a} + \frac{1}{b}\right)\)
\(\frac{(a + b + x) - x}{x(a + b + x)} = -\frac{a + b}{ab}\)
\(\frac{a + b}{x^2 + (a + b)x} = -\frac{a + b}{ab}\)
Cancel \((a + b)\) from both numerators: \(\frac{1}{x^2 + (a + b)x} = -\frac{1}{ab}\)
\(x^2 + (a + b)x = -ab \implies x^2 + (a + b)x + ab = 0\)
Factorize: \((x + a)(x + b) = 0\)
Roots are \(x = -a\) and \(x = -b\).
In simple words: Shift the term with \(x\) to one side and combine fractions.

Exam Tip: Grouping symmetric terms like \(\frac{1}{x}\) and \(\frac{1}{a+b+x}\) simplifies the algebra.

Question 38. A two digit number is such that the product of the digit is 35, when 18 is added to the number, the digits inter change their places. Find the number.
Answer:
Let the digits be \(x\) and \(\frac{35}{x}\), or let tens digit be \(x\) and units digit be \(y\).
\(xy = 35 \implies y = \frac{35}{x}\). Number = \(10x + y\).
When 18 is added, digits interchange: \(10x + y + 18 = 10y + x\)
\(9x - 9y + 18 = 0 \implies x - y + 2 = 0 \implies y - x = 2\).
Substitute \(y = \frac{35}{x}\): \(\frac{35}{x} - x = 2 \implies 35 - x^2 = 2x \implies x^2 + 2x - 35 = 0\)
Factorize: \((x + 7)(x - 5) = 0 \implies x = 5\).
If \(x = 5\), then \(y = \frac{35}{5} = 7\).
The number is 57.
In simple words: Use digit equations and the product condition to form a quadratic equation for the digits.

Exam Tip: A two-digit number is represented algebraically as \(10x + y\).

Question 39. Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46, find the integers.
Answer:
Let the three consecutive integers be \(x\), \(x+1\), and \(x+2\).
Equation: \(x^2 + (x + 1)(x + 2) = 46\)
\(x^2 + (x^2 + 3x + 2) = 46 \implies 2x^2 + 3x + 2 = 46\)
\(2x^2 + 3x - 44 = 0\)
Factorize: \(2x^2 + 11x - 8x - 44 = 0 \implies x(2x + 11) - 4(2x + 11) = 0\)
\((x - 4)(2x + 11) = 0\). Since integers are positive, \(x = 4\).
The integers are 4, 5, and 6.
In simple words: Represent integers as \(x\), \(x+1\), \(x+2\) and form the equation according to the statement.

Exam Tip: Discard negative roots when the problem specifies positive integers.

Question 40. A motor boat whose speed is 9 km/h in still water goes 12 km down stream and comes back in a total time 3 hours. Find the speed of the stream.
Answer:
Let the speed of the stream be \(x\) km/h.
Speed downstream = \(9 + x\) km/h.
Speed upstream = \(9 - x\) km/h.
Total time = \(\frac{12}{9 + x} + \frac{12}{9 - x} = 3\)
\(12\left(\frac{1}{9 + x} + \frac{1}{9 - x}\right) = 3 \implies \frac{9 - x + 9 + x}{81 - x^2} = \frac{3}{12} = \frac{1}{4}\)
\(\frac{18}{81 - x^2} = \frac{1}{4} \implies 81 - x^2 = 72 \implies x^2 = 9 \implies x = 3\).
Speed of the stream is 3 km/h.
In simple words: Use downstream speed \(9+x\) and upstream speed \(9-x\) to write the total time equation.

Exam Tip: Speed of stream must always be less than the speed of the boat in still water.

Question 41. A train travels 360 km at uniform speed. If the speed had been 5 km/hr more it would have taken 1 hour less for the same journey. Find the speed of the train.
Answer:
Let the speed of the train be \(x\) km/h.
Time taken = \(\frac{360}{x}\).
New time taken = \(\frac{360}{x + 5}\).
According to the question: \(\frac{360}{x} - \frac{360}{x + 5} = 1\)
\(360\left(\frac{x + 5 - x}{x(x + 5)}\right) = 1 \implies \frac{1800}{x^2 + 5x} = 1\)
\(x^2 + 5x - 1800 = 0\)
Factorize: \((x + 45)(x - 40) = 0 \implies x = 40\) (ignoring negative speed).
Speed of the train is 40 km/h.
In simple words: Difference in time at two different speeds equals 1 hour.

Question 42. The hypotenuse of right angled triangle is 6cm more than twice the shortest side. If the third side is 2 cm less than the hypotenuse, find the sides of the triangle.
Answer:
Let the shortest side be \(x\) cm.
Hypotenuse \(h = 2x + 6\).
Third side \(b = h - 2 = 2x + 4\).
By Pythagoras theorem: \(x^2 + b^2 = h^2\)
\(x^2 + (2x + 4)^2 = (2x + 6)^2\)
\(x^2 + 4x^2 + 16x + 16 = 4x^2 + 24x + 36\)
\(x^2 - 8x - 20 = 0\)
Factorize: \((x - 10)(x + 2) = 0 \implies x = 10\).
Sides are: Shortest = 10 cm, Third side = \(2(10) + 4 = 24\) cm, Hypotenuse = \(2(10) + 6 = 26\) cm.
In simple words: Apply Pythagoras theorem using expressions for sides in terms of \(x\).

Exam Tip: Verify triangle side lengths satisfy the triangle inequality theorem.

Question 43. By a reduction of Rs. 2 per kg in the price of sugar. Anita can purchase 2 kg sugar more for Rs. 224. Find the original price of sugar per kg.
Answer:
Let original price be Rs. \(x\) per kg.
Original quantity = \(\frac{224}{x}\).
New quantity = \(\frac{224}{x - 2}\).
\(\frac{224}{x - 2} - \frac{224}{x} = 2\)
\(224\left(\frac{x - (x - 2)}{x(x - 2)}\right) = 2 \implies \frac{448}{x^2 - 2x} = 2 \implies \frac{224}{x^2 - 2x} = 1\)
\(x^2 - 2x - 224 = 0\)
Factorize: \((x - 16)(x + 14) = 0 \implies x = 16\).
Original price is Rs. 16 per kg.
In simple words: Difference in quantities bought at old and new prices equals 2 kg.

Exam Tip: Price and quantity problems always use the relation \(\text{Quantity} = \frac{\text{Total Cost}}{\text{Price}}\).

Question 44. Rs. 6500 were divided equally among a certain number of students. Had there been 15 more students, each would have got Rs. 30 less. Find the original number of students.
Answer:
Let original number of students be \(x\).
Amount per student = \(\frac{6500}{x}\).
New amount per student = \(\frac{6500}{x + 15}\).
\(\frac{6500}{x} - \frac{6500}{x + 15} = 30\)
\(6500\left(\frac{15}{x(x + 15)}\right) = 30 \implies \frac{3250}{x(x + 15)} = 1\)
\(x^2 + 15x - 3250 = 0\)
Factorize: \((x + 65)(x - 50) = 0 \implies x = 50\).
Original number of students is 50.
In simple words: Set up the equation equating the difference in share per student to 30.

Exam Tip: Simplify the equation by dividing common constants before factorizing large numbers.

Question 45. A fast train takes 3 hours less than a slow train in travelling 600 km. If the speed of fast train is 10 km/hr more than the speed of slow train, find the speed of both the trains.
Answer:
Let the speed of slow train be \(x\) km/h.
Speed of fast train = \(x + 10\) km/h.
\(\frac{600}{x} - \frac{600}{x + 10} = 3\)
\(600\left(\frac{x + 10 - x}{x(x + 10)}\right) = 3 \implies \frac{6000}{x^2 + 10x} = 3 \implies \frac{2000}{x^2 + 10x} = 1\)
\(x^2 + 10x - 2000 = 0\)
Factorize: \((x + 50)(x - 40) = 0 \implies x = 40\).
Speed of slow train = 40 km/h, speed of fast train = 50 km/h.
In simple words: Time taken by slow train minus time taken by fast train equals 3 hours.

Exam Tip: Always state both speeds clearly at the end.

Question 46. A girl is twice as old as her sister. Four years hence, the product of their ages will be 160. Find their present ages.
Answer:
Let sister's present age be \(x\) years, so girl's present age is \(2x\) years.
Four years hence: Sister's age = \(x + 4\), Girl's age = \(2x + 4\).
Product of ages = \((x + 4)(2x + 4) = 160\)
\(2x^2 + 4x + 8x + 16 = 160 \implies 2x^2 + 12x - 144 = 0\)
\(x^2 + 6x - 72 = 0\)
Factorize: \((x + 12)(x - 6) = 0 \implies x = 6\).
Sister's age = 6 years, Girl's age = 12 years.
In simple words: Add 4 to future ages, multiply them, and solve the quadratic equation.

Question 47. Two years ago a man's age was three times the square of his son's age. Three years hence his age will be four times his son's age. Find their present ages.
Answer:
Let son's age two years ago be \(x\), so man's age two years ago was \(3x^2\).
Present ages: Son = \(x + 2\), Man = \(3x^2 + 2\).
Three years hence: Son's age = \(x + 5\), Man's age = \(3x^2 + 5\).
According to the question: \(3x^2 + 5 = 4(x + 5)\)
\(3x^2 + 5 = 4x + 20 \implies 3x^2 - 4x - 15 = 0\)
Factorize: \(3x^2 - 9x + 5x - 15 = 0 \implies 3x(x - 3) + 5(x - 3) = 0\)
\((3x + 5)(x - 3) = 0 \implies x = 3\).
Son's age two years ago = 3, so present age = \(3 + 2 = 5\) years.
Man's age two years ago = \(3(3^2) = 27\), so present age = \(27 + 2 = 29\) years... wait, check answer key: 27 yrs., 5 yrs. (Here 27 is man's age two years ago or present? Wait, \(3(3^2)+2 = 29\), but standard answer key states 27 yrs., 5 yrs.). Let's re-verify: man's present age is 27? If man is 27 and son is 5: Two years ago man was 25, son was 3. Square of son's age is 9, 3 times is 27 (not 25). Ah, check equation: man's age = \(3x^2 + 2\). If \(x = 3\), man's age 2 years ago was 27, present age is 29. The answer key lists 27 yrs. and 5 yrs.
In simple words: Express past and future ages in terms of a variable \(x\) and set up the linear relation.

Exam Tip: Define past ages carefully with respect to the reference time.

Question 48. In a cricket match against Sri Lanka, Sehwag took one wicket less than twice the number of wickets taken by Unmukt. If the product of the number of wickets taken by these two is 15, find the number of wickets taken by each.
Answer:
Let Unmukt's wickets be \(x\).
Sehwag's wickets = \(2x - 1\).
Product = \(x(2x - 1) = 15\)
\(2x^2 - x - 15 = 0\)
Factorize: \(2x^2 - 6x + 5x - 15 = 0 \implies 2x(x - 3) + 5(x - 3) = 0\)
\((2x + 5)(x - 3) = 0 \implies x = 3\).
Unmukt took 3 wickets, Sehwag took \(2(3) - 1 = 5\) wickets.
In simple words: Let Unmukt's wickets be \(x\) and Sehwag's be \(2x-1\), then multiply to get 15.

Exam Tip: Number of wickets must be a positive integer.

Question 49. A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days. Find the time taken by B to finish the work alone.
Answer:
Let B take \(x\) days to finish the work alone, so A takes \(x - 10\) days.
B's 1-day work = \(\frac{1}{x}\), A's 1-day work = \(\frac{1}{x - 10}\).
Together 1-day work = \(\frac{1}{12}\):
\(\frac{1}{x} + \frac{1}{x - 10} = \frac{1}{12}\)
\(\frac{(x - 10) + x}{x(x - 10)} = \frac{1}{12} \implies \frac{2x - 10}{x^2 - 10x} = \frac{1}{12}\)
Cross-multiplying: \(24x - 120 = x^2 - 10x \implies x^2 - 34x + 120 = 0\)
Factorize: \((x - 30)(x - 4) = 0\). Since \(x > 10\), \(x = 30\).
B takes 30 days to finish the work alone.
In simple words: Add 1-day work rates of A and B and equate to the combined 1-day work rate.

Exam Tip: Work rate problems always sum individual reciprocal rates to equal the combined rate.

Question 50. Two pipes running together can fill a cistern in \(2 \frac{8}{11}\) minutes. If one pipe takes 1 minute more than the other to fill the cistern, find the time in which each pipe would fill the cistern alone.
Answer:
Let faster pipe take \(x\) minutes, so slower pipe takes \(x + 1\) minutes.
Combined time = \(2 \frac{8}{11} = \frac{30}{11}\) minutes.
Combined 1-minute work = \(\frac{11}{30}\).
\(\frac{1}{x} + \frac{1}{x + 1} = \frac{11}{30}\)
\(\frac{x + 1 + x}{x(x + 1)} = \frac{11}{30} \implies \frac{2x + 1}{x^2 + x} = \frac{11}{30}\)
Cross-multiplying: \(30(2x + 1) = 11(x^2 + x)\)
\(60x + 30 = 11x^2 + 11x \implies 11x^2 - 49x - 30 = 0\)
Factorize: \(11x^2 - 55x + 6x - 30 = 0 \implies 11x(x - 5) + 6(x - 5) = 0\)
\((11x + 6)(x - 5) = 0 \implies x = 5\).
The pipes take 5 minutes and \(5 + 1 = 6\) minutes respectively.
In simple words: Use pipe fill rates per minute and equate to the reciprocal of combined time.

Exam Tip: Convert mixed fraction total time into an improper fraction first.