CBSE Class 10 Maths HOTs Quadratic Equations Set 08

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 \(x = -\left(\frac{ac + bc - 2ab}{a + b - 2c}\right)\)