CBSE Class 10 Maths HOTs Quadratic Equations Set H

Very Short Answer Type Questions

Question. Find the nature of the roots of the following quadratic equation. If the real roots exist, find them : \( 3x^2 - 4\sqrt{3}x + 4 = 0 \)
Answer: We have \( 3x^2 - 4\sqrt{3}x + 4 = 0 \)
Compare with \( ax^2 + bx + c = 0 \) we get
\( a = 3, b = -4\sqrt{3}, c = 4 \)
\( b^2 - 4ac = (-4\sqrt{3})^2 - 4(3)(4) \)
\( = 48 - 48 = 0 \)
Thus roots are real and equal.
Roots are \( (-\frac{b}{2a}), (-\frac{b}{2a}) \) or \( \frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3} \)

Question. If the equation \( kx^2 - 2kx + 6 = 0 \) has equal roots, then find the value of \( k \).
Answer: We have \( kx^2 - 2kx + 6 = 0 \)
Compare with \( ax^2 + bx + c = 0 \) we get
\( a = k, b = -2k, c = 6 \)
Since roots of the equation are equal, \( b^2 - 4ac = 0 \)
\( (-2k)^2 - 4(k)(6) = 0 \)
\( 4k^2 - 24k = 0 \)
\( 4k(k - 6) = 0 \)
\( k = 0, 6 \)
But \( k \neq 0 \), as coefficient of \( x^2 \) can not be zero
\( k = 6 \)

Question. Find the values of \( p \) for which the quadratic equation \( 4x^2 + px + 3 = 0 \) has equal roots.
Answer: We have \( 4x^2 + px + 3 = 0 \)
Compare with \( ax^2 + bx + c = 0 \) we get
\( a = 4, b = p, c = 3 \)
Since roots of the equation are equal, \( b^2 - 4ac = 0 \)
\( p^2 - 4 \times 4 \times 3 = 0 \)
\( p^2 - 48 = 0 \)
\( p^2 = 48 \)
\( p = \pm 4\sqrt{3} \)

Question. A train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.
Answer: Let the usual speed of train be \(x \text{ km/hr}\).
According to question we have
\(\frac{300}{x} - \frac{300}{x + 5} = 2\)
\(x^2 + 5x - 750 = 0\)
\(x^2 + 30x - 25x - 750 = 0\)
\((x + 30)(x - 25) = 0\)
\(x = -30\) or \(x = 25\)
Since, speed cannot be negative \(x \neq -30\).
Thus speed of train is 25 km/hr.

Question. Solve, for \(x\) : \(\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0\)
Answer: We have \(\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0\)
\(\sqrt{3}x^2 + 3x + 7x + 7\sqrt{3} = 0\)
\(\sqrt{3}x(x + \sqrt{3}) + 7(x + \sqrt{3}) = 0\)
\((x + \sqrt{3})(\sqrt{3}x + 7) = 0\)
\(x = -\sqrt{3}\) and \(x = \frac{-7}{\sqrt{3}}\)
Hence roots \(x = -\sqrt{3}\) or \(x = \frac{-7}{\sqrt{3}}\)

Question. Find the nature of the roots of the quadratic equation : \( 13\sqrt{3}x^2 + 10x + \sqrt{3} = 0 \)
Answer: We have \( 13\sqrt{3}x^2 + 10x + \sqrt{3} = 0 \)
Compare with \( ax^2 + bx + c = 0 \) we get
\( a = 13\sqrt{3}, b = 10, c = \sqrt{3} \)
\( b^2 - 4ac = (10)^2 - 4(13\sqrt{3})(\sqrt{3}) \)
\( = 100 - 156 \)
\( = -56 \)
As \( D < 0 \), the equation has not real roots.

Question. Determine the positive value of 'k' for which the equation \( x^2 + kx + 64 = 0 \) and \( x^2 - 8x + k = 0 \) will both have real and equal roots.
Answer: We have \( x^2 + kx + 64 = 0 \)
Compare with \( ax^2 + bx + c = 0 \) we get
\( a = 1, b = k, c = 64 \)
For real and equal roots, \( b^2 - 4ac = 0 \)
Thus \( k^2 - 4 \times 1 \times 64 = 0 \)
\( k^2 - 256 = 0 \)
\( k = \pm 16 \)
(1)

Question. Find the roots of the quadratic equation \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)
Answer: We have \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)
\(\sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0\)
\(\sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0\)
\((x + \sqrt{2})(\sqrt{2}x + 5) = 0\)
Thus \(x = -\sqrt{2}\) and \(x = -\frac{5}{\sqrt{2}} = -\frac{5}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = -\frac{5}{2}\sqrt{2}\)

Question. Find the value of \(k\) for which the roots of the quadratic equation \(2x^2 + kx + 8 = 0\) will have the equal roots ?
Answer: We have \(2x^2 + kx + 8 = 0\)
Compare with \(ax^2 + bx + c = 0\) we get \(a = 2, b = k\), and \(c = 8\)
For equal roots, \(D = 0\)
\(b^2 - 4ac = 0\)
\(k^2 - 4 \times 2 \times 8 = 0\)
\(k^2 = 64\)
\(k = \pm\sqrt{64}\)
Thus \(k = \pm 8\)

Question. Solve for \(x : \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0\)
Answer: We have \(\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0\)
\(\sqrt{3}x^2 + 3x + 7x + 7\sqrt{3} = 0\)
\(\sqrt{3}x(x + \sqrt{3}) + 7(x + \sqrt{3}) = 0\)
\((x + \sqrt{3})(\sqrt{3}x + 7) = 0\)
Thus \(x = -\sqrt{3}\) and \(x = -\frac{7}{\sqrt{3}}\)

Question. Find the roots of the quadratic equation : \( a^2b^2x^2 + b^2x - a^2x - 1 = 0 \)
Answer: We have \( a^2b^2x^2 + b^2x - a^2x - 1 = 0 \)
\( b^2x(a^2x + 1) - 1(a^2x + 1) = 0 \)
\( (b^2x - 1)(a^2x + 1) = 0 \)
\( x = \frac{1}{b^2} \) or \( x = -\frac{1}{a^2} \)
Hence, roots are \( \frac{1}{b^2} \) and \( -\frac{1}{a^2} \).

Short Answer Type Questions

Question. Solve the following quadratic equation for \( x \):
\( 9x^2 - 6b^2x - (a^4 - b^4) = 0 \)
Answer: We have \( 9x^2 - 6b^2x - (a^4 - b^4) = 0 \)
Compare with \( ax^2 + bx + c = 0 \) we have
\( a = 9, b = -6b^2, c = -(a^4 - b^4) \)
\( x = \frac{6b^2 \pm \sqrt{(-6b^2)^2 - 4 \times 9 \times \{-(a^4 - b^4)\}}}{2 \times 9} \)
\( = \frac{6b^2 \pm \sqrt{36b^4 + 36a^4 - 36b^4}}{18} \)
\( = \frac{6b^2 \pm \sqrt{36a^4}}{18} \)
\( = \frac{6b^2 \pm 6a^2}{18} \)
Thus \( x = \frac{a^2 + b^2}{3}, \frac{b^2 - a^2}{3} \)

Question. Solve the quadratic equation, \( 2x^2 + ax - a^2 = 0 \) for \( x \).
Answer: We have \( 2x^2 + ax - a^2 = 0 \)
Compare with \( Ax^2 + Bx + C = 0 \) we have
\( A = 2, B = a, C = -a^2 \)
Now \( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)
\( = \frac{-a \pm \sqrt{a^2 - 4 \times 2 \times (-a^2)}}{2 \times 2} \)
\( = \frac{-a \pm \sqrt{a^2 + 8a^2}}{4} \)
\( = \frac{-a \pm \sqrt{9a^2}}{4} \)
\( = \frac{-a \pm 3a}{4} \)
\( x = \frac{-a + 3a}{4}, \frac{-a - 3a}{4} \)
Thus \( x = \frac{a}{2}, -a \)

Question. Find the roots of the quadratic equation \( 4x^2 - 4px + (p^2 - q^2) = 0 \)
Answer: We have \( 4x^2 - 4px + (p^2 - q^2) = 0 \)
Compare with \( ax^2 + bx + c = 0 \) we get
\( a = 4, b = -4p, c = (p^2 - q^2) \)
The roots are given by the quadratic formula
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{4p \pm \sqrt{16p^2 - 4 \times 4 \times (p^2 - q^2)}}{2 \times 4} \)
\( = \frac{4p \pm \sqrt{16p^2 - 16p^2 + 16q^2}}{8} \)
\( = \frac{4p \pm 4q}{8} \)
Thus roots are \( \frac{p+q}{2}, \frac{p-q}{2} \).

Question. Find the value of \(k\), for which one root of the quadratic equation \(kx^2 - 14x + 8 = 0\) is six times the other.
Answer: We have \(kx^2 - 14x + 8 = 0\)
Let one root be \(\alpha\) and other root be \(6\alpha\).
Sum of roots \(\alpha + 6\alpha = \frac{14}{k}\)
\(7\alpha = \frac{14}{k}\) or \(\alpha = \frac{2}{k}\) ...(1)
Product of roots \(\alpha(6\alpha) = \frac{8}{k}\)
or, \(6\alpha^2 = \frac{8}{k}\) ...(2)
Solving (1) and (2), we obtain
\(6(\frac{2}{k})^2 = \frac{8}{k}\)
\(6 \times \frac{4}{k^2} = \frac{8}{k}\)
\(\frac{3}{k^2} = \frac{1}{k}\)
\(3k = k^2\)
\(3k - k^2 = 0\)
\(k(3 - k) = 0\)
\(k = 0\) or \(k = 3\)
Since \(k = 0\) is not possible, therefore \(k = 3\).

Question. Find the value(s) of \(k\) if the quadratic equation \(3x^2 - k\sqrt{3}x + 4 = 0\) has real roots.
Answer: If discriminant of quadratic equation is equal to zero, or more than zero, then roots are real.
We have \(3x^2 - k\sqrt{3}x + 4 = 0\)
Compare with \(ax^2 + bx + c = 0\)
\(D = b^2 - 4ac\)
For real roots \(b^2 - 4ac \ge 0\)
\((-k\sqrt{3})^2 - 4 \times 3 \times 4 \ge 0\)
\(3k^2 - 48 \ge 0\)
\(k^2 - 16 \ge 0\)
\((k - 4)(k + 4) \ge 0\)
Thus \(k \le -4\) and \(k \ge 4\)

Question. Sum of the areas of two squares is \( 468 \text{ m}^2 \). If the difference of their perimeter is \( 24 \text{ m} \), find the sides of the squares.
Answer: Let the side of the smaller square be \( y \) and be the side of the longer square be \( x \), then we have
\( 4x - 4y = 24 \)
\( x - y = 6 \)
\( x = y + 6 \)
According to the question
\( x^2 + y^2 = 468 \)
\( (y + 6)^2 + y^2 = 468 \)
\( 2y^2 + 12y + 36 = 468 \)
\( 2y^2 + 12y - 432 = 0 \)
\( y^2 + 6y - 216 = 0 \)
\( (y + 18)(y - 12) = 0 \)
\( y = -18, 12 \)
As side can not be negative, \( y = 12 \) and \( x = 12 + 6 = 18 \)
Hence, the side of larger square \( 18 \text{ m} \) and that of smaller square \( 12 \text{ m} \).

Question. Solve for \( x \) (in terms of \( a \) and \( b \)):
\( \frac{a}{x - b} + \frac{b}{x - a} = 2, x \neq a, b \)
Answer: We have \( \frac{a(x - a) + b(x - b)}{(x - b)(x - a)} = 2 \)
\( a(x - a) + b(x - b) = 2[x^2 - (a + b)x + ab] \)
\( ax - a^2 + bx - b^2 = 2x^2 - 2(a + b)x + 2ab \)
\( 2x^2 - 3(a + b)x + (a + b)^2 = 0 \)
\( 2x^2 - 2(a + b)x - (a + b)x + (a + b)^2 = 0 \)
\( [2x - (a + b)][x - (a + b)] = 0 \)
Thus \( x = a + b, \frac{a + b}{2} \)

Question. If \( x = \frac{2}{3} \) and \( x = -3 \) are roots of the quadratic equation \( ax^2 + 7x + b = 0 \), find the values of \( a \) and \( b \).
Answer: We have \( ax^2 + 7x + b = 0 \)
Substituting \( x = \frac{2}{3} \) in above equation we obtain
\( \frac{4}{9}a + \frac{14}{3} + b = 0 \)
\( 4a + 42 + 9b = 0 \)
\( 4a + 9b = -42 \) ...(1)
and substituting \( x = -3 \) we obtain
\( 9a - 21 + b = 0 \)
\( 9a + b = 21 \) ...(2)
Solving (1) and (2), we get \( a = 3 \) and \( b = -6 \)

Question. Solve for \( x \): \( \sqrt{6x + 7} - (2x - 7) = 0 \)
Answer: We have \( \sqrt{6x + 7} - (2x - 7) = 0 \)
or, \( \sqrt{6x + 7} = (2x - 7) \)
Squaring both sides we get
\( 6x + 7 = (2x - 7)^2 \)
\( 6x + 7 = 4x^2 - 28x + 49 \)
\( 4x^2 - 34x + 42 = 0 \)
\( 2x^2 - 17x + 21 = 0 \)
\( 2x^2 - 14x - 3x + 21 = 0 \)
\( 2x(x - 7) - 3(x - 7) = 0 \)
\( (x - 7)(2x - 3) = 0 \)
Thus \( x = 7 \) and \( x = \frac{3}{2} \)

Question. A two digit number is four times the sum of the digits. It also equal to 3 times the product of digits. Find the number.
Answer: Let units digit and tens digit of the two digit number be \( x \) and \( y \) respectively.
Thus number is \( 10y + x \)
According to question, we have
\( 10y + x = 4(y + x) \)
\( 10y + x = 4y + 4x \)
\( 10y - 4y = 4x - x \)
\( 6y = 3x \)
\( 2y = x \)
Also, \( 10y + x = 3xy \)
\( 10y + 2y = 3(2y)y \)
\( 12y = 6y^2 \)
\( 6y^2 - 12y = 0 \)
\( 6y(y - 2) = 0 \)
\( y = 0 \text{ or } y = 2 \)
As the number can not be zero \( x = 4 \) and \( x = 2y = 4 \).
Thus required number is 24.

Question. Solve for \(x : \sqrt{2x+9} + x = 13\)
Answer: We have \(\sqrt{2x+9} + x = 13\)
\(\sqrt{2x+9} = 13 - x\)
Squaring both side we have
\(2x+9 = (13 - x)^2\)
\(2x+9 = 169 + x^2 - 26x\)
\(0 = x^2 + 169 - 26x - 9 - 2x\)
\(x^2 - 28x + 160 = 0\)
\(x^2 - 20x - 8x + 160 = 0\)
\(x(x - 20) - 8(x - 20) = 0\)
\((x - 8)(x - 20) = 0\)
Thus \(x = 8\) and \(x = 20\).

Question. Solve the equation for \( x : \frac{4}{x} - 3 = \frac{5}{2x + 3} ; x \neq 0, \frac{-3}{2} \)
Answer: We have \( \frac{4}{x} - 3 = \frac{5}{2x + 3} \)
\( \frac{4}{x} - \frac{5}{2x + 3} = 3 \)
\( \frac{4(2x + 3) - 5x}{x(2x + 3)} = 3 \)
\( \frac{8x + 12 - 5x}{x(2x + 3)} = 3 \)
\( 3x + 12 = 3x(2x + 3) \)
\( 3x + 12 = 6x^2 + 9x \)
\( 6x^2 + 6x - 12 = 0 \)
\( x^2 + x - 2 = 0 \)
\( x^2 + 2x - x - 2 = 0 \)
\( x(x + 2) - (x + 2) = 0 \)
\( (x + 2)(x - 1) = 0 \)
Thus \( x = -2, 1 \)

Question. Find the roots of the equation \( 2x^2 + x - 4 = 0 \), by the method of completing the squares.
Answer: We have \( 2x^2 + x - 4 = 0 \)
\( x^2 + \frac{x}{2} - 2 = 0 \)
\( x^2 + 2x\left(\frac{1}{4}\right) - 2 = 0 \)
Adding and subtracting \( \left(\frac{1}{4}\right)^2 \), we get
\( x^2 + 2x\left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2 - 2 = 0 \)
\( \left(x + \frac{1}{4}\right)^2 - \left(\frac{1}{16} + 2\right) = 0 \)
\( \left(x + \frac{1}{4}\right)^2 - \left(\frac{1 + 32}{16}\right) = 0 \)
\( \left(x + \frac{1}{4}\right)^2 - \frac{33}{16} = 0 \)
\( \left(x + \frac{1}{4}\right)^2 = \frac{33}{16} \)
\( x + \frac{1}{4} = \pm \frac{\sqrt{33}}{4} \)
The roots are \( x = \frac{-1 \pm \sqrt{33}}{4} \), i.e., \( \left(\frac{-1 + \sqrt{33}}{4}\right) \) and \( \left(\frac{-1 - \sqrt{33}}{4}\right) \)

Question. Solve for \( x : 9x^2 - 6ax + (a^2 - b^2) = 0 \)
Answer: We have \( 9x^2 - 6ax + a^2 - b^2 = 0 \)
Compare with \( Ax^2 + Bx + C = 0 \) we have
\( A = 9, B = -6a, C = (a^2 - b^2) \)
\( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)
\( x = \frac{6a \pm \sqrt{(-6a)^2 - 4 \times 9(a^2 - b^2)}}{2 \times 9} \)
\( x = \frac{6a \pm \sqrt{36a^2 - 36a^2 + 36b^2}}{18} \)
\( x = \frac{6a \pm 6b}{18} \)
\( x = \frac{6(a + b)}{18}, x = \frac{6(a - b)}{18} \)
\( x = \frac{a + b}{3}, x = \frac{a - b}{3} \)

Question. Solve for \(x : x^2 + 5x - (a^2 + a - 6) = 0\)
Answer: \(x^2 + 5x - (a^2 + a - 6) = 0\)
\(x = \frac{-5 \pm \sqrt{25 + 4(a^2 + a - 6)}}{2}\)
\(\because x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(x = \frac{-5 \pm \sqrt{25 + 4a^2 + 4a - 24}}{2}\)
\(x = \frac{-5 \pm \sqrt{4a^2 + 4a + 1}}{2}\)
\(x = \frac{-5 \pm (2a + 1)}{2}\)
\(x = \frac{2a - 4}{2}, \frac{-2a - 6}{2}\)
\(x = a - 2, x = -(a + 3)\)

Question. Solve for \(x : x^2 - (2b - 1)x + (b^2 - b - 20) = 0\)
Answer: We have \(x^2 - (2b - 1)x + (b^2 - b - 20) = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we have
\(A = 1, B = -(2b - 1), C = (b^2 - b - 20)\)
\(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\)
\(x = \frac{(2b - 1) \pm \sqrt{(2b - 1)^2 - 4(b^2 - b - 20)}}{2}\)
\(x = \frac{(2b - 1) \pm \sqrt{4b^2 - 4b + 1 - 4b^2 + 4b + 80}}{2}\)
\(x = \frac{(2b - 1) \pm 9}{2}\)
\(x = \frac{2b + 8}{2}, \frac{2b - 10}{2}\)
\(x = b + 4, b - 5\)
Thus \(x = b + 4\) and \(x = b - 5\)

Question. Solve the equation \( \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}, x \neq -4, 7 \) for \( x \).
Answer: We have, \( \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30} \)
\( \frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30} \)
\( \frac{x - 7 - x - 4}{(x + 4)(x - 7)} = \frac{11}{30} \)
\( \frac{-11}{(x + 4)(x - 7)} = \frac{11}{30} \)
\( \frac{-1}{(x + 4)(x - 7)} = \frac{1}{30} \)
\( (x + 4)(x - 7) = -30 \)
\( x^2 - 3x - 28 = -30 \)
\( x^2 - 3x + 2 = 0 \)
\( x^2 - 2x - x + 2 = 0 \)
\( (x - 1)(x - 2) = 0 \)
Thus \( x = 1, 2 \).

Question. Solve the following quadratic equation for \( x : p^2x^2 + (p^2 - q^2)x - q^2 = 0 \)
Answer: We have \( p^2x^2 + (p^2 - q^2)x - q^2 = 0 \)
Compare with \( ax^2 + bx + c = 0 \) we get
\( a = p^2, b = p^2 - q^2, c = -q^2 \)
The roots are given by the quadratic formula
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-(p^2 - q^2) \pm \sqrt{(p^2 - q^2)^2 - 4(p^2)(-q^2)}}{2p^2} \)
\( x = \frac{-(p^2 - q^2) \pm \sqrt{p^4 + q^4 - 2p^2q^2 + 4p^2q^2}}{2p^2} \)
\( x = \frac{-(p^2 - q^2) \pm \sqrt{p^4 + q^4 + 2p^2q^2}}{2p^2} \)
\( x = \frac{-(p^2 - q^2) \pm \sqrt{(p^2 + q^2)^2}}{2p^2} \)
\( x = \frac{-(p^2 - q^2) \pm (p^2 + q^2)}{2p^2} \)
Thus \( x = \frac{-(p^2 - q^2) + (p^2 + q^2)}{2p^2} = \frac{2q^2}{2p^2} = \frac{q^2}{p^2} \)
and \( x = \frac{-(p^2 - q^2) - (p^2 + q^2)}{2p^2} = \frac{-2p^2}{2p^2} = -1 \)
Hence, roots are \( \frac{q^2}{p^2}, -1 \)

Question. Solve the following quadratic equation for \( x : 9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0 \)
Answer: We have \( 9x^2 - 9(a + b)x + 2a^2 + 5ab + 2b^2 = 0 \)
Now \( 2a^2 + 5ab + 2b^2 = 2a^2 + 4ab + ab + 2b^2 \)
\( = 2a(a + 2b) + b(a + 2b) \)
\( = (a + 2b)(2a + b) \)
Hence the equation becomes
\( 9x^2 - 9(a + b)x + (a + 2b)(2a + b) = 0 \)
\( 9x^2 - 3[3(a + b)]x + (a + 2b)(2a + b) = 0 \)
\( 9x^2 - 3[(a + 2b) + (2a + b)]x + (a + 2b)(2a + b) = 0 \)
\( 9x^2 - 3(a + 2b)x - 3(2a + b)x + (a + 2b)(2a + b) = 0 \)
\( 3x[3x - (a + 2b)] - (2a + b)[3x - (a + 2b)] = 0 \)
\( [3x - (a + 2b)][3x - (2a + b)] = 0 \)
\( 3x - (a + 2b) = 0 \Rightarrow x = \frac{a + 2b}{3} \)
\( 3x - (2a + b) = 0 \Rightarrow x = \frac{2a + b}{3} \)
Hence, roots are \( \frac{a + 2b}{3} \) and \( \frac{2a + b}{3} \).

Question. Solve for \( x : x^2 + 6x - (a^2 + 2a - 8) = 0 \)
Answer: We have \( x^2 + 6x - (a^2 + 2a - 8) = 0 \)
Compare with \( Ax^2 + Bx + C = 0 \) we get
\( A = 1, B = 6, C = -(a^2 + 2a - 8) \)
The roots are given by the quadratic formula
\( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)
\( x = \frac{-6 \pm \sqrt{36 + 4(a^2 + 2a - 8)}}{2} \)
\( x = \frac{-6 \pm \sqrt{36 + 4a^2 + 8a - 32}}{2} \)
\( x = \frac{-6 \pm \sqrt{4a^2 + 8a + 4}}{2} \)
\( x = \frac{-6 \pm \sqrt{(2a + 2)^2}}{2} \)
\( x = \frac{-6 \pm (2a + 2)}{2} \)
Thus \( x = \frac{-6 + (2a + 2)}{2} = \frac{2a - 4}{2} = a - 2 \)
and \( x = \frac{-6 - (2a + 2)}{2} = \frac{-2a - 8}{2} = -a - 4 \)
Thus \( x = a - 2, -a - 4 \)

Question. Three consecutive natural number are such that the squre of the middle number exceeds the difference of the squares of the other two by 60. Find the number.
Answer: Let the three consecutive natural numbers be \(x, x+1\) and \(x+2\).
Now \((x+1)^2 = (x+2)^2 - x^2 + 60\)
\(x^2 + 2x + 1 = x^2 + 4x + 4 - x^2 + 60\)
\(x^2 - 2x - 63 = 0\)
\(x^2 - 9x + 7x - 63 = 0\)
\(x(x - 9) + 7(x - 9) = 0\)
\((x - 9)(x + 7) = 0\)
\(x = 9\) or \(x = -7\)
As \(x\) can’t be negative, \(x = 9\).
Hence three numbers are 9, 10, 11.

Question. The sum of ages (in years) of a son and his father is 35 years and product of theirs ages is 150 years, find their ages.
Answer: Let the age of father be \(x\) years and age of son be \(y\) years
Now \(x + y = 35\) (1)
and \(xy = 150\) (2)
Putting the value of \(y\), from (1) we have
\(x(35 - x) = 150\)
\(x^2 - 35x + 150 = 0\)
\((x - 30)(x - 5) = 0\)
\(x = 30, x = 5\) (Rejected)
Age of father cant be 5 years, so we reject \(x = 5\) and take \(x = 30\).
Now \(y = 5\)
Hence the age of father is 30 years and the age of son is 5 years.

Question. One fourth of a herd of camels was seen in forest. Twice of square root of the herd had gone to mountain and remaining 15 camels were seen on the bank of a river, find the total number of camels.
Answer: Let the total number of camels be \(x\).
According to the question,
\(\frac{x}{4} + 2\sqrt{x} + 15 = x\)
\(3x - 8\sqrt{x} - 60 = 0\)
Let \(\sqrt{x} = y\), then we have
\(3y^2 - 8y - 60 = 0\)
\(3y^2 - 18y + 10y - 60 = 0\)
\(3y(y - 6) + 10(y - 6) = 0\)
\((3y + 10)(y - 6) = 0\)
\(y = 6\) or \(y = -\frac{10}{3}\)
Here \(y = -\frac{10}{3}\) is not possible.
Thus \(y = 6\) or \(y^2 = 36\), \(x = y^2 = 36\)
Hence the number of camels is 36.

Question. The sum of the squares of two consecutive naturals is 421. Find the numbers.
Answer: Let the first natural number be \(x\). The second consecutive natural will be \(x + 1\)
According to the question,
\(x^2 + (x + 1)^2 = 421\)
\(x^2 + x^2 + 2x + 1 = 421\)
\(x^2 + x - 210 = 0\)
\(x^2 + 15x - 14x - 210 = 0\)
\(x(x + 15) - 14(x + 15) = 0\)
\((x + 15)(x - 14) = 0\)
\(x + 15 = 0\) or \(x - 14 = 0\)
\(x = -15\) or \(x = 14\)
Rejecting negative value \(x = 14\).
Therefore first number is 14 and consecutive number is 15.

Question. In a class test, the sum of the marks obtained by a student in mathematics and science is 28. Had he got 3 marks more in mathematics and 4 marks less in science, the product of the marks would have been 180. Find his marks in two subjects.
Answer: Let marks obtained in maths be \(x\), the marks obtained in science will be \(28 - x\)
Now \((x + 3)(28 - x - 4) = 180\)
\((x + 3)(24 - x) = 180\)
\(24x - x^2 + 72 - 3x = 180\)
\(x^2 - 21x + 108 = 0\)
\((x - 9)(x - 12) = 0\)
\(x = 9\) or \(x = 12\)
Case I : \(x = 9\)
Marks obtained in maths = 9
Marks obtained in science = \(28 - 9 = 19\)
Case II : \(x = 12\)
Marks obtained in maths = 12
Marks obtained in science = \(28 - 12 = 16\)

Question. If the roots of the equation \((a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0\) are equal, prove that \(\frac{a}{b} = \frac{c}{d}\).
Answer: We have \((a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0\)
Compare with \(Ax^2 + Bx + C = 0\) we get
\(A = (a^2 + b^2), B = -2(ac + bd), C = (c^2 + d^2)\)
If roots are equal, \(D = B^2 - 4AC = 0\)
or \(B^2 = 4AC\)
Now \([-2(ac + bd)]^2 = 4(a^2 + b^2)(c^2 + d^2)\)
\((a^2c^2 + 2abcd + b^2d^2) = (a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2)\)
\(2abcd = a^2d^2 + b^2c^2\)
\(0 = a^2d^2 - 2abcd + b^2c^2\)
\(0 = (ad - bc)^2\)
\(0 = ad - bc\)
\(ad = bc\)
\(\frac{a}{b} = \frac{c}{d}\) Hence Proved

Question. Solve for \(x : \left(\frac{2x}{x-5}\right)^2 + 5\left(\frac{2x}{x-5}\right) - 24 = 0, x \neq 5\)
Answer: We have \(\left(\frac{2x}{x-5}\right)^2 + 5\left(\frac{2x}{x-5}\right) - 24 = 0\)
Let \(\frac{2x}{x-5} = y\) then we have
\(y^2 + 5y - 24 = 0\)
\((y + 8)(y - 3) = 0\)
\(y = 3, -8\)
Taking \(y = 3\) we have
\(\frac{2x}{x-5} = 3\)
\(2x = 3x - 15\)
\(x = 15\)
Taking \(y = -8\) we have
\(\frac{2x}{x-5} = -8\)
\(2x = -8x + 40\)
\(10x = 40\)
\(x = 4\)
Hence, \(x = 15, 4\)

Question. Solve for \(x : \frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4}, x \neq -1, -2, -4\)
Answer: We have \(\frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4}\)
\(\frac{(x+2) + 2(x+1)}{(x+1)(x+2)} = \frac{4}{x+4}\)
\(\frac{x+2+2x+2}{x^2+3x+2} = \frac{4}{x+4}\)
\(\frac{3x+4}{x^2+3x+2} = \frac{4}{x+4}\)
\((3x+4)(x+4) = 4(x^2+3x+2)\)
\(3x^2+16x+16 = 4x^2+12x+8\)
\(x^2 - 4x - 8 = 0\)
Now \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(= \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-8)}}{2 \times 1}\)
\(= \frac{4 \pm \sqrt{16+32}}{2}\)
\(= \frac{4 \pm \sqrt{48}}{2} = \frac{4 \pm 4\sqrt{3}}{2}\)
\(= 2 \pm 2\sqrt{3}\)
Hence, \(x = 2 + 2\sqrt{3}\) and \(2 - 2\sqrt{3}\)

Question. Find \(x\) in terms of \(a, b\) and \(c : \frac{a}{x-a} + \frac{b}{x-b} = \frac{2c}{x-c}, x \neq a, b, c\)
Answer: We have \(\frac{a}{x-a} + \frac{b}{x-b} = \frac{2c}{x-c}\)
\(a(x-b)(x-c) + b(x-a)(x-c) = 2c(x-a)(x-b)\)
\(ax^2 - abx - acx + abc + bx^2 - bax - bcx + abc = 2cx^2 - 2cxb - 2cxa + 2abc\)
\(ax^2 + bx^2 - 2cx^2 - abx - acx - bax - bcx + 2cbx + 2acx = 0\)
\(x^2(a + b - 2c) - 2abx + acx + bcx = 0\)
\(x^2(a + b - 2c) + x(ac + bc - 2ab) = 0\)
Thus \(x = -\left(\frac{ac + bc - 2ab}{a + b - 2c}\right)\)