CBSE Class 10 Maths HOTs Quadratic Equations Set B

IMPORTANT TERMS, DEFINITIONS AND FORMULAE

• The polynomial of degree two is called quadratic polynomial and equation corresponding to a quadratic polynomial \( P(x) \) is called a quadratic equation in variable \( x \).
  Thus, \( P(x) = ax^2 + bx + c = 0, a \neq 0, a, b, c \in \mathbb{R} \) is known as the standard form of quadratic equation.
• Roots/solutions of a quadratic equation: The values of \( x \) that satisfy an equation are called the solution or roots of the equation.
  A real number \( m \) is said to be a solution/root of the quadratic equation \( ax^2 + bx + c = 0 \), if \( am^2 + bm + c = 0 \).
• The graph of a quadratic equation is parabolic in shape.
• Methods of solving a quadratic equation:
  A quadratic equation of the form \( ax^2 + bx + c = 0 \), can be solved by any of the following methods:
  (a) Factorization : Splitting the middle term method. (b) Quadratic Formula: We can calculate the roots by using \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

(i) Form the product “\( ac \)”.

(ii) Find a pair of numbers \( b_1 \) and \( b_2 \) whose product is “\( ac \)” and whose sum is “\( b \)” (if you can’t find such number, it can’t be factorized).

(iii) Split the middle term using \( b_1 \) and \( b_2 \), that expresses the term \( bx \) as \( b_1x \) and \( b_2x \). Now factor by grouping the pairs of terms.

(iv) Roots of the quadratic equation can be found by equating each linear factor to zero. Since product of two numbers is zero, so either or both of them are zero.

• The expression \( b^2 - 4ac \) is called the discriminant of the quadratic equation \( ax^2 + bx + c = 0 \). It is denoted by \( D \).
• Nature of Roots

(i) If \( D = b^2 - 4ac > 0 \), then the roots are real and distinct.

(ii) If \( D = b^2 - 4ac = 0 \), the roots are real and equal or coincident.

(iii) If \( D = b^2 - 4ac < 0 \), the roots are not real (imaginary roots).

• If \( \alpha \) and \( \beta \) are two roots of an equation then the required quadratic equation can be formed as \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \) or \( x^2 - (\alpha + \beta)x + \alpha\beta = 0 \).
• Let \( \alpha \) and \( \beta \) be two roots of the quadratic equation \( ax^2 + bx + c = 0 \), then
  Sum of Roots = \( (-\text{the coefficient of } x) / (\text{the coefficient of } x^2) \Rightarrow \alpha + \beta = \frac{-b}{a} \)
  Product of Roots = \( \text{constant term} / \text{the coefficient of } x^2 \Rightarrow \alpha\beta = \frac{c}{a} \)

SHORT ANSWER TYPE QUESTIONS – I

Question. If \( x = 3 \) is one root of the quadratic equation \( x^2 - 2mx - 6 = 0 \), then find the value of \( m \).
Answer: Given equation is : \( x^2 - 2mx - 6 = 0 \)
\( \because x = 3 \) is a root of the given equation
\( \Rightarrow (3)^2 - 2m(3) - 6 = 0 \Rightarrow 9 - 6m - 6 = 0 \Rightarrow 3 - 6m = 0 \)
\( \Rightarrow 6m = 3 \) or \( m = \frac{1}{2} \)

Question. Find the nature of roots of the quadratic equation \( 4x^2 + 4\sqrt{3}x + 3 = 0 \).
Answer: Given equation is \( 4x^2 + 4\sqrt{3}x + 3 = 0 \)
Comparing it with \( ax^2 + bx + c = 0 \)
\( a = 4, b = 4\sqrt{3}, c = 3 \)
\( D = b^2 - 4ac = (4\sqrt{3})^2 - 4(4)(3) = 48 - 48 = 0 \)
As \( D = 0 \), the equation has real and equal roots.

Question. If \( \alpha \) and \( \beta \) are the roots of \( ax^2 - bx + c = 0 \), \( (a \neq 0) \), then find the value of \( \alpha + \beta \).
Answer: We know that \( \alpha + \beta = \frac{-B}{A} \)
Here, \( A = a, B = -b, C = c \)
\( \therefore \alpha + \beta = \frac{-(-b)}{a} = \frac{b}{a} \)

Question. Form a quadratic equation whose roots are \( \frac{-1}{2} \) and \( \frac{5}{3} \).
Answer: We know that, the quadratic equation can be formed by
\( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \)
So, required equation is \( x^2 - (\frac{-1}{2} + \frac{5}{3})x + (\frac{-1}{2} \times \frac{5}{3}) = 0 \)
\( \Rightarrow x^2 - \frac{7}{6}x + \frac{-5}{6} = 0 \)

Question. Is there any real value of 'k' for which the equation \( x^2 + 2x + (k^2 + 1) = 0 \) has real roots?
Answer: Given equation, \( x^2 + 2x + (k^2 + 1) = 0 \)
For real roots \( D = b^2 - 4ac \geq 0 \)
Here, \( a = 1, b = 2, c = k^2 + 1 \)
\( \therefore D = (2)^2 - 4(1)(k^2 + 1) \geq 0 \)
\( \Rightarrow 4 - 4k^2 - 4 \geq 0 \)
\( \Rightarrow -4k^2 \geq 0 \), which is not possible for any real value of \( k \) (except \( k = 0 \) for which \( D = 0 \)).
So, no real value of \( k \) exists for which the given equation has real roots (assuming \( D > 0 \) as per text).

Question. Is \( x^2 - 5\sqrt{x} + 7 = 0 \) a quadratic equation? Justify your answer.
Answer: No, because \( x^2 - 5\sqrt{x} + 7 \) is not a quadratic polynomial, as it contains a term involving \( x^{1/2} \), where \( \frac{1}{2} \) is not an integer.

Question. If \( x = 1 \) is a common root of the equations \( ax^2 + ax + 4 = 0 \) and \( x^2 + x + b = 0 \), then find \( ab \).
Answer: Given, \( x = 1 \) is a root of \( ax^2 + ax + 4 = 0 \)
\( \Rightarrow a + a + 4 = 0 \Rightarrow a = -2 \)
Also, \( x = 1 \) is a root of \( x^2 + x + b = 0 \)
\( \Rightarrow 1 + 1 + b = 0 \Rightarrow b = -2 \)
\( \therefore ab = (-2)(-2) = 4 \).

Question. If the sum of the roots of the equation \( x^2 - x - \lambda(2x - 1) = 0 \) is zero, then find \( \lambda \).
Answer: Given, \( x^2 - x - 2\lambda x + \lambda = 0 \)
\( \Rightarrow x^2 - (2\lambda + 1)x + \lambda = 0 \)
Comparing it with \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \)
\( \Rightarrow \text{Sum of roots} = 2\lambda + 1 = 0 \Rightarrow \lambda = -\frac{1}{2} \)

Question. If \( \alpha \) and \( \beta \) are the roots of \( x^2 - 5x + p = 0 \) and \( \alpha - \beta = 1 \), then find \( p \).
Answer: Given, \( x^2 - 5x + p = 0 \)
Comparing it with \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \)
\( \Rightarrow \alpha + \beta = 5 \) and \( \alpha\beta = p \)
Also, \( \alpha - \beta = 1 \) (given)
On solving, \( \alpha = 3, \beta = 2 \)
\( \therefore p = \text{product of roots} = (\alpha\beta) = 6 \).

Question. Solve \( 4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0 \) using factorisation method.
Answer: Given \( 4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0 \)
\( \Rightarrow 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0 \)
\( \Rightarrow 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0 \)
\( \Rightarrow (4x - \sqrt{3})(\sqrt{3}x + 2) = 0 \Rightarrow x = \frac{\sqrt{3}}{4} \) or \( x = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \).

Question. Find the discriminant of \( 3x^2 - 2\sqrt{6}x - 2 = 0 \).
Answer: Given, \( 3x^2 - 2\sqrt{6}x - 2 = 0 \)
Here \( a = 3, b = -2\sqrt{6}, c = -2 \)
\( D = b^2 - 4ac = (-2\sqrt{6})^2 - 4(3)(-2) = 24 + 24 = 48 \).

Question. For what values of \( k \), the roots of the equation \( 3x^2 - 10x + k = 0 \) are reciprocal of each other.
Answer: Given, \( 3x^2 - 10x + k = 0 \).
Let roots be \( \alpha \) and \( \beta = \frac{1}{\alpha} \), then
\( \alpha\beta = \alpha \times \frac{1}{\alpha} = \frac{k}{3} \Rightarrow 1 = \frac{k}{3} \Rightarrow k = 3 \).

Question. If the quadratic equation \( px^2 - 2\sqrt{5}px + 15 = 0 \) has two real and equal roots, then find the value of \( p \).
Answer: Given, \( px^2 - 2\sqrt{5}px + 15 = 0 \)
Here, \( a = p, b = -2\sqrt{5}p \) and \( c = 15 \)
For real and equal roots, \( D = 0 \) or \( b^2 - 4ac = 0 \)
\( \Rightarrow (-2\sqrt{5}p)^2 - 4(p)(15) = 0 \Rightarrow 20p^2 - 60p = 0 \)
\( \Rightarrow p^2 - 3p = 0 \Rightarrow p(p - 3) = 0 \Rightarrow p = 3 \) or \( p = 0 \).
So, \( p = 3 \) as for \( p = 0 \), quadratic equation is not possible.

Question. Find the roots of the quadratic equation \( \sqrt{3}x^2 - 2x - \sqrt{3} = 0 \).
Answer: Given, \( \sqrt{3}x^2 - 2x - \sqrt{3} = 0 \)
Comparing it with \( ax^2 + bx + c = 0 \), here, \( a = \sqrt{3}, b = -2, c = -\sqrt{3} \).
\( \therefore x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(\sqrt{3})(-\sqrt{3})}}{2\sqrt{3}} = \frac{2 \pm \sqrt{16}}{2\sqrt{3}} = \frac{2 \pm 4}{2\sqrt{3}} \)
\( \Rightarrow x = \frac{6}{2\sqrt{3}} \) or \( x = \frac{-2}{2\sqrt{3}} \).
\( \Rightarrow x = \sqrt{3} \) or \( x = -\frac{\sqrt{3}}{3} \).

Question. For what value of \( k, kx^2 + 4x + 1 = 0 \) has real roots?
Answer: Here, \( a = k, b = 4, c = 1 \)
\( D = b^2 - 4ac = 16 - 4 \times k \times 1 = 16 - 4k \)
For real roots \( D \geq 0 \Rightarrow 16 - 4k \geq 0 \Rightarrow k \leq 4 \).

Question. Find the roots of the quadratic equation \( \frac{1}{2}x^2 - \sqrt{11}x + 1 = 0 \), by using the quadratic formula.
Answer: Given, \( \frac{1}{2}x^2 - \sqrt{11}x + 1 = 0 \)
Here \( a = \frac{1}{2}, b = -\sqrt{11}, c = 1 \)
\( \Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \Rightarrow x = \frac{\sqrt{11} \pm \sqrt{11 - 2}}{1} \Rightarrow x = \sqrt{11} \pm 3 \).

Question. If \( -5 \) is a root of the quadratic equation \( 2x^2 + px - 15 = 0 \) and the quadratic equation \( p(x^2 + x) + k = 0 \) has equal roots, then find \( k \).
Answer: Given, \( -5 \) is a root of \( 2x^2 + px - 15 = 0 \)
\( \Rightarrow 2(-5)^2 + p(-5) - 15 = 0 \)
\( \Rightarrow 50 - 5p - 15 = 0 \Rightarrow p = 7 \)
\( \therefore \) The other equation is \( 7x^2 + 7x + k = 0 \)
It has equal roots, so \( D = 0 \).
\( \Rightarrow b^2 - 4ac = 0 \Rightarrow 49 - 28k = 0 \Rightarrow k = \frac{7}{4} \).

Question. Solve for \( x : \frac{1}{(x - 3)} - \frac{1}{(x + 5)} = \frac{1}{6}, x \neq 3, -5
Answer: \( \frac{1}{(x - 3)} - \frac{1}{(x + 5)} = \frac{1}{6} \)
\( \Rightarrow \frac{(x + 5) - (x - 3)}{(x - 3)(x + 5)} = \frac{1}{6} \Rightarrow 48 = x^2 + 2x - 15 \)
\( \Rightarrow x^2 + 2x - 63 = 0 \Rightarrow x^2 + 9x - 7x - 63 = 0 \)
\( \Rightarrow (x + 9)(x - 7) = 0 \) or \( x = 7, -9 \).

Question. Two positive numbers differ by 3 and their product is 504. Find the numbers.
Answer: Let one number be \( x \), then the other number is \( (x + 3) \).
ATQ, \( x(x + 3) = 504 \)
\( \Rightarrow x^2 + 3x - 504 = 0 \Rightarrow x^2 + 24x - 21x - 504 = 0 \Rightarrow x(x + 24) - 21(x + 24) = 0 \)
\( \Rightarrow (x - 21)(x + 24) = 0 \Rightarrow x = -24 \) or \( x = 21 \), \( x \) cannot be negative
\( \therefore x = 21 \) and \( x + 3 = 24 \).

Question. Find the value of \( k \) for which the equation \( x^2 + k(2x + k - 1) = 0 \), has real and equal roots.
Answer: Let \( x^2 + k(2x + k - 1) = 0 \Rightarrow x^2 + 2kx + k^2 - k = 0 \)
Here, \( a = 1, b = 2k, c = k^2 - k \)
Since, roots are real and equal, \( D = b^2 - 4ac = 0 \)
\( \Rightarrow (2k)^2 - 4(1)(k^2 - k) = 0 \Rightarrow 4k = 0 \Rightarrow k = \frac{0}{4} = 0 \).

Question. If one root of the quadratic equation \( 2x^2 - 3x + p = 0 \) is 3, find the other root of the quadratic equation. Also, find the value of \( p \).
Answer: Given, \( x = 3 \) is a root of \( 2x^2 - 3x + p = 0 \)
\( \Rightarrow 2(3)^2 - 3(3) + p = 0 \Rightarrow 18 - 9 + p = 0 \Rightarrow 9 + p = 0 \Rightarrow p = -9 \).
\( \therefore \) Equation is \( 2x^2 - 3x - 9 = 0 \)
\( \Rightarrow 2x^2 - 6x + 3x - 9 = 0 \Rightarrow 2x(x - 3) + 3(x - 3) = 0 \)
\( \Rightarrow (x - 3)(2x + 3) = 0 \Rightarrow x = 3 \) or \( x = -\frac{3}{2} \).
Hence, other root is \( -\frac{3}{2} \).

Question. If \( x = \frac{2}{3} \) and \( x = -3 \) are roots of the equation \( ax^2 + 7x + b = 0 \), find \( a \) and \( b \).
Answer: Given, \( x = \frac{2}{3} \) and \( x = -3 \) are roots of the equation \( ax^2 + 7x + b = 0 \).
\( \Rightarrow a(\frac{2}{3})^2 + 7(\frac{2}{3}) + b = 0 \)
\( \Rightarrow 4a + 42 + 9b = 0 \) ...(i)
Also, \( a(-3)^2 + 7(-3) + b = 0 \)
\( \Rightarrow 9a - 21 + b = 0 \) ...(ii)
On solving (i) and (ii) simultaneously, we get \( a = 3, b = -6 \).

Question. Solve the quadratic equation \( 2x^2 + ax - a^2 = 0 \) for \( x \).
Answer: The given equation is \( 2x^2 + ax - a^2 = 0 \).
By quadratic formula \( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)
\( \Rightarrow x = \frac{-a \pm \sqrt{a^2 + 8a^2}}{4} \) [Here \( B = a, A = 2, C = -a^2 \)]
\( \Rightarrow x = \frac{-a \pm 3a}{4} \Rightarrow x = \frac{-a-3a}{4} \) or \( x = \frac{-a+3a}{4} \)
\( \Rightarrow x = -a \) or \( x = \frac{a}{2} \)

Question. Find the values of \( p \) so that the quadratic equation \( 3x^2 - 2px + 12 = 0 \) has equal roots.
Answer: For equal roots, \( D = b^2 - 4ac = 0 \)
Here, \( a = 3, b = -2p, c = 12 \)
\( \therefore D = b^2 - 4ac = (-2p)^2 - 4(3)(12) = 0 \)
\( \Rightarrow 4p^2 - 144 = 0 \Rightarrow (2p + 12)(2p - 12) = 0 \Rightarrow p = -6 \) or \( p = 6 \)
\( \therefore p = \pm 6 \)

Question. 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 part be \( x \). Then smaller part = \( 16 - x \)
ATQ, \( 2x^2 = (16 - x)^2 + 164 \)
\( \Rightarrow 2x^2 - (16 - x)^2 - 164 = 0 \Rightarrow x^2 + 32x - 420 = 0 \)
\( \Rightarrow (x + 42)(x - 10) = 0 \Rightarrow x = -42 \) or \( x = 10 \)
\( \therefore x = 10 \)
So, the required parts are 10 and 6.

Question. Solve for \( x : 4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0 \)
Answer: Here, \( a = 4\sqrt{3}, b = 5, c = -2\sqrt{3} \)
So, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \Rightarrow x = \frac{-5 \pm \sqrt{25 + 96}}{8\sqrt{3}} \)
\( \Rightarrow x = \frac{-5 \pm \sqrt{121}}{8\sqrt{3}} \Rightarrow x = \frac{-5 + 11}{8\sqrt{3}} \) or \( x = \frac{-5 - 11}{8\sqrt{3}} \)
\( \Rightarrow x = \frac{3}{4\sqrt{3}} = \frac{\sqrt{3}}{4} \) or \( x = \frac{-2}{\sqrt{3}} = \frac{-2\sqrt{3}}{3} \).

Question. The sum of a number and its positive square root is \( \frac{6}{25} \). Find the number.
Answer: Let the number be \( x^2 \) and its positive square root be \( x \).
ATQ, \( x^2 + x = \frac{6}{25} \)
\( \Rightarrow 25x^2 + 25x - 6 = 0 \Rightarrow 25x^2 + 30x - 5x - 6 = 0 \)
\( \Rightarrow 5x(5x + 6) - 1(5x + 6) = 0 \)
\( \Rightarrow (5x - 1)(5x + 6) = 0 \)
\( \Rightarrow x = \frac{1}{5} \) or \( x = \frac{-6}{5} \)
Since, \( x \) is positive, so \( x = \frac{1}{5} \)
\( \therefore \) The number is \( x^2 = \frac{1}{25} \).

Question. Solve \( x^2 + x + 1 = 0 \) using quadratic formula.
Answer: Given, \( x^2 + x + 1 = 0 \)
Here, \( a = 1, b = 1, c = 1 \)
\( \therefore D = b^2 - 4ac \Rightarrow D = 1 - 4 = -3 \)
\( \Rightarrow D < 0 \).
\( \therefore \) The given quadratic equation has no solution.

Question. Solve : \( \frac{4}{x} - 3 = \frac{5}{2x+3}; x \neq 0, \frac{-3}{2} \), for \( x \).
Answer: Given, \( \frac{4}{x} - 3 = \frac{5}{2x+3} \)
\( \Rightarrow (4 - 3x)(2x + 3) = 5x \Rightarrow 6x^2 + 6x - 12 = 0 \Rightarrow x^2 + x - 2 = 0 \)
\( \Rightarrow x(x + 2) - 1(x + 2) = 0 \Rightarrow (x - 1)(x + 2) = 0 \Rightarrow x = -2 \) or 1

Question. Find the values of \( a \) and \( b \), if the sum and the product of the roots of the equation \( 4ax^2 + 4bx + 3 = 0 \) are \( \frac{1}{2} \) and \( \frac{3}{16} \) respectively.
Answer: Given, \( 4ax^2 + 4bx + 3 = 0 \)
Let \( \alpha \) and \( \beta \) be its roots.
Here, \( A = 4a, B = 4b, C = 3 \)
\( \alpha + \beta = \frac{-B}{A} = \frac{-4b}{4a} = -\frac{b}{a} \)
\( \Rightarrow -\frac{b}{a} = \frac{1}{2} \) or \( -2b = a \) ...(i)
Product of roots = \( \frac{C}{A} \)
\( \Rightarrow \frac{3}{16} = \frac{3}{4a} \Rightarrow a = 4 \) ...(ii)
\( \therefore -2b = a \Rightarrow b = -2 \)
\( \therefore a = 4 \) and \( b = -2 \).

SHORT ANSWER TYPE QUESTIONS– II 

Question. The sum of the squares of two consecutive natural numbers is 421, find the numbers.
Answer: Let the numbers be \( x \) and \( (x + 1) \).
ATQ, \( x^2 + (x + 1)^2 = 421 \)
\( \Rightarrow x^2 + x^2 + 2x + 1 = 421 \Rightarrow 2x^2 + 2x - 420 = 0 \)
\( \Rightarrow x^2 + x - 210 = 0 \Rightarrow x^2 + 15x - 14x - 210 = 0 \)
\( \Rightarrow x(x + 15) - 14(x + 15) = 0 \)
\( \Rightarrow (x - 14)(x + 15) = 0 \Rightarrow x = 14, -15 \)
\( x = -15 \) is rejected as it is not a natural number.
\( \therefore x = 14 \) and \( x + 1 = 15 \)
Hence, the numbers are 14 and 15.

Question. If the roots of \( x^2(a^2 + b^2) + 2(ac + bd)x + (c^2 + d^2) = 0 \) are equal, then prove that \( ad = bc \).
Answer: Given, \( x^2(a^2 + b^2) + 2(ac + bd)x + (c^2 + d^2) = 0 \)
For equal roots \( D = B^2 - 4AC = 0 \)
\( \Rightarrow [2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0 \)
\( \Rightarrow 4a^2c^2 + 4b^2d^2 + 8acbd - 4a^2c^2 - 4a^2d^2 - 4b^2c^2 - 4b^2d^2 = 0 \)
\( \Rightarrow 8abcd - 4a^2d^2 - 4b^2c^2 = 0 \)
\( \Rightarrow 4[2abcd - a^2d^2 - b^2c^2] = 0 \)
\( \Rightarrow -4[a^2d^2 + b^2c^2 - 2abcd] = 0 \)
\( \Rightarrow -4[ad - bc]^2 = 0 \)
\( \Rightarrow ad - bc = 0 \)
\( \Rightarrow ad = bc \). Proved.

Question. Solve for \( x : x^2 + 5x - (a^2 + a - 6) = 0 \).
Answer: Given equation, \( x^2 + 5x - (a^2 + 3a - 2a - 6) = 0 \)
\( \Rightarrow x^2 + 5x - (a - 2)(a + 3) = 0 \)
\( \Rightarrow x^2 + [(a + 3) - (a - 2)]x - (a - 2)(a + 3) = 0 \)
\( \Rightarrow x^2 + (a + 3)x - (a - 2)x - (a - 2)(a + 3) = 0 \)
\( \Rightarrow x[x + (a + 3)] - (a - 2)[x + (a + 3)] = 0 \)
\( \Rightarrow [x - (a - 2)][x + (a + 3)] = 0 \)
\( \Rightarrow x = a - 2 \) or \( x = -(a + 3) \).

Question. Solve for \( x : 9x^2 - 6ax + (a^2 - b^2) = 0 \).
Answer: Given, \( 9x^2 - 6ax + (a^2 - b^2) = 0 \Rightarrow (9x^2 - 6ax + a^2) - b^2 = 0 \)
\( \Rightarrow [(3x)^2 - 2(3x)(a) + (a)^2] - b^2 = 0 \)
\( \Rightarrow [3x - a]^2 - b^2 = 0 \Rightarrow (3x - a - b)(3x - a + b) = 0 \)
\( \Rightarrow 3x - a - b = 0 \) or \( 3x - a + b = 0 \)
\( \Rightarrow x = \frac{a+b}{3} \) or \( x = \frac{a-b}{3} \)

Question. The sum of ages of a son and his father is 35 years and product of their ages is 150. Find their ages.
Answer: Let the age of father be \( x \).
Then age of son will be \( 35 - x \)
ATQ, \( x(35 - x) = 150 \)
\( \Rightarrow 35x - x^2 = 150 \Rightarrow x^2 - 35x + 150 = 0 \)
\( \Rightarrow x^2 - 30x - 5x + 150 = 0 \Rightarrow x(x - 30) - 5(x - 30) = 0 \)
\( \Rightarrow (x - 5)(x - 30) = 0 \Rightarrow x = 5 \) or \( x = 30 \)
\( x = 5 \) will be rejected as \( 35 - x = 30 \) cannot be age of son.
\( x = 30 \)
Hence, age of father is 30 years and the age of son = \( 35 - x = 5 \) years.

Question. For what value of \( k \), are the roots of the quadratic equation, \( (k + 4)x^2 + (k + 1)x + 1 = 0 \) equal?
Answer: Given, \( (k + 4)x^2 + (k + 1)x + 1 = 0 \)
Here \( a = (k + 4), b = (k + 1), c = 1 \)
For equal roots, \( D = b^2 - 4ac = 0 \)
\( \Rightarrow (k + 1)^2 - 4(k + 4)(1) = 0 \Rightarrow k^2 + 1 + 2k - 4k - 16 = 0 \)
\( \Rightarrow k^2 - 2k - 15 = 0 \Rightarrow k^2 - 5k + 3k - 15 = 0 \)
\( \Rightarrow k(k - 5) + 3(k - 5) = 0 \Rightarrow (k + 3)(k - 5) = 0 \)
\( \Rightarrow k = -3 \) or \( k = 5 \).

Question. If \( (x^2 + y^2)(a^2 + b^2) = (ax + by)^2 \), prove that \( \frac{b}{a} = \frac{y}{x} \).
Answer: Given, \( (x^2 + y^2)(a^2 + b^2) = (ax + by)^2 \)
\( \Rightarrow a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 = a^2x^2 + b^2y^2 + 2axby \)
\( \Rightarrow a^2y^2 + b^2x^2 - 2abxy = 0 \Rightarrow (ay - bx)^2 = 0 \Rightarrow ay - bx = 0 \)
\( \Rightarrow \frac{y}{x} = \frac{b}{a} \) or \( \frac{b}{a} = \frac{y}{x} \). Proved.

Question. If the quadratic equation \( (1 + m^2)n^2x^2 + 2mnpx + (p^2 - s^2) = 0 \) in \( x \) has equal roots, show that \( p^2 = s^2(1 + m^2) \).
Answer: Given, \( (1 + m^2)n^2x^2 + 2mnpx + (p^2 - s^2) = 0 \)
Here, \( a = (1 + m^2)n^2, b = 2mnp, c = p^2 - s^2 \).
For equal roots : \( D = b^2 - 4ac = 0 \)
\( \Rightarrow 4m^2n^2p^2 - 4(1 + m^2)n^2(p^2 - s^2) = 0 \)
\( \Rightarrow 4m^2n^2p^2 - 4n^2(p^2 + p^2m^2 - s^2 - s^2m^2) = 0 \)
\( \Rightarrow 4n^2[m^2p^2 - (p^2 + m^2p^2 - s^2 - s^2m^2)] = 0 \)
\( \Rightarrow m^2p^2 - p^2 - m^2p^2 + s^2 + s^2m^2 = 0 \Rightarrow -p^2 + s^2(1 + m^2) = 0 \)
\( \Rightarrow p^2 = s^2(1 + m^2) \). Proved.

Question. The length of the sides forming right angle of a right triangle are \( 5x \) cm and \( (3x - 1) \) cm. If the area of the triangle is \( 60 \text{ cm}^2 \). find its hypotenuse.
Answer: Let the base and perpendicular of the right angle triangle be \( (3x - 1) \) and \( 5x \).
Given, \( \text{area} = 60 \text{ cm}^2 \)
We know that, \( \text{area of triangle} = \frac{1}{2} \times b \times h = A \)
\( \Rightarrow \frac{1}{2} \times (3x - 1)(5x) = 60 \)
\( \Rightarrow 15x^2 - 5x - 120 = 0 \Rightarrow 3x^2 - x - 24 = 0 \)
\( \Rightarrow 3x^2 - 9x + 8x - 24 = 0 \Rightarrow 3x(x - 3) + 8(x - 3) = 0 \)
\( \Rightarrow (x - 3)(3x + 8) = 0 \Rightarrow x = 3 \) or \( x = -\frac{3}{8} \) (rejected)
\( \therefore x = 3 \)
Hence, \( \text{base} = 3x - 1 = 8 \text{ cm} \) and \( \text{perpendicular} = 5x = 15 \text{ cm} \)
\( \therefore \text{Hypotenuse} = \sqrt{8^2 + 15^2} = 17 \text{ cm} \)

Question. Determine the positive value of \( k \) for which the equations \( x^2 + kx + 64 = 0 \) and \( x^2 - 8x + k = 0 \) will have both real and equal roots.
Answer: Given, equation \( x^2 + 4x + 64 = 0 \) has real and equal roots.
So, \( b^2 - 4ac = 0 \Rightarrow k^2 - 4(64) = 0 \)
\( \Rightarrow k^2 = 256 \) or \( k = \pm 16 \) ...(i)
Also, given \( x^2 - 8x + k = 0 \) has real and equal roots
\( \therefore b^2 - 4ac = 0 \)
\( \Rightarrow (-8)^2 - 4(1)(k) = 0 \Rightarrow 64 - 4k = 0 \Rightarrow k = 16 \) ...(ii)
\( \therefore \) From (i) and (ii) the positive value of \( k \) will be 16.