CBSE Class 10 Mathematics Quadratic Equations VBQs Set J

Question. For what value of \( k \) is one root of the quadratic equation \( 9x^2 - 18x + k = 0 \) double the other?
(a) 36
(b) 9
(c) 12
(d) 8
Answer: (d) 8
Explanation :
Let one root be \( \alpha \) and another root be \( 2\alpha \)
So,
\( \alpha + 2\alpha = \frac{-(-18)}{9} = 2 \)
\( 3\alpha = 2 \)
\( \Rightarrow \alpha = \frac{2}{3} \)
Now \( \alpha \times 2\alpha = \frac{k}{9} \)
\( 2\alpha^2 = \frac{k}{9} \)
\( \Rightarrow 2 \times \frac{4}{9} = \frac{k}{9} \)
\( \Rightarrow k = 8 \)

Question. The condition for the sum and the product of the roots of the quadratic equation \( ax^2 - bx + c = 0 \) to be equal, is
(a) \( b + c = 0 \)
(b) \( b - c = 0 \)
(c) \( a + c = 0 \)
(d) \( a + b + c = 0 \)
Answer: (b) b – c = 0
Explanation :
\( \therefore \) Sum of roots = Product of roots
\( \frac{b}{a} = \frac{c}{a} \)
\( \Rightarrow b = c \)
\( \Rightarrow b - c = 0 \)

Question. Find the value of \( \sqrt{30 + \sqrt{30 + \sqrt{30 + \dots \infty}}} \)
(a) 6
(b) – 5
(c) Either (1) and (2)
(d) Neither (1) nor (2)
Answer: (a) 6
Explanation :
Let \( y = \sqrt{30 + \sqrt{30 + \sqrt{30 + \dots \infty}}} \)
\( y = \sqrt{30 + y} \)
Squaring both sides
\( y^2 = 30 + y \)
\( y^2 - y - 30 = 0 \)
\( y^2 - 6y + 5y - 30 = 0 \)
\( y(y - 6) + 5(y - 6) = 0 \)
\( (y + 5)(y - 6) = 0 \)
\( y = - 5 \) and \( y = 6 \)
But \( x \) must be positive
\( x = 6 \)

Question. The roots of \( x^2 - (a + 1)x + b^2 = 0 \) are equal. Then choose the correct value of \( a, b \) from the following option :
(a) 5, 2
(b) 3, 4
(c) 5, – 3
(d) 5, 4
Answer: (c) 5, – 3
Explanation :
The roots of \( x^2 - (a + 1)x + b^2 = 0 \) are equal
\( \Rightarrow (a + 1)^2 - 4b^2 = 0 \)
\( \Rightarrow a + 1 = \pm 2b \)
From the options
\( a = 5, b = - 3 \) satifes the above relation.

Question. The solution of the equation \( x^2 + x + 1 = 1 \) is
(a) \( x = 0 \)
(b) \( x = - 1 \)
(c) Both (a) and (b)
(d) Cannot be determined
Answer: (c) Both (a) and (b)
Explanation :
Given equation is
\( x^2 + x + 1 = 1 \)
\( x^2 + x = 0 \)
\( x(x + 1) = 0 \)
\( x = 0 \) and \( x = - 1 \)

Question. If one of the roots of an equation, \( x^2 - 2x + c = 0 \) is thrice the other, then \( c = ? \)
(a) \( \frac{1}{2} \)
(b) \( \frac{4}{3} \)
(c) \( -\frac{1}{2} \)
(d) \( \frac{3}{4} \)
Answer: (d) \( \frac{3}{4} \)
Explanation :
Let one root be \( \alpha \) and another root be \( 3\alpha \)
So,
\( \alpha + 3\alpha = \frac{-(-2)}{1} = 2 \)
\( 4\alpha = 2 \)
\( \Rightarrow \alpha = \frac{1}{2} \)
Now \( \alpha \times 3\alpha = c \)
\( 3\alpha^2 = c \)
\( 3 \times \frac{1}{4} = c \)
\( \Rightarrow c = \frac{3}{4} \)

Question. The roots of the equation \( x^2 + 5x + 1 = 0 \) are
(a) \( \frac{5 \pm \sqrt{21}}{2}, \frac{5 - \sqrt{21}}{2} \)
(b) \( \frac{-5 - \sqrt{21}}{2}, \frac{5 + \sqrt{21}}{2} \)
(c) \( \frac{-5 + \sqrt{21}}{2}, \frac{-5 - \sqrt{21}}{2} \)
(d) \( \frac{-5 + \sqrt{29}}{2}, \frac{-5 - \sqrt{29}}{2} \)
Answer: (c) \( \frac{-5 + \sqrt{21}}{2}, \frac{-5 - \sqrt{21}}{2} \)
Explanation :
Given equation
\( x^2 + 5x + 1 = 0 \)
By quadratic formula
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-5 \pm \sqrt{25 - 4}}{2} \)
\( x = \frac{-5 \pm \sqrt{21}}{2} \)

Question. The roots of \( x^2 - 2x - 1 = 0 \) are ..............
(a) \( \sqrt{2} + 1, \sqrt{2} - 1 \)
(b) \( 1, \sqrt{2} \)
(c) \( 1 + \sqrt{2}, 1 - \sqrt{2} \)
(d) 2, 1
Answer: (c) \( 1 + \sqrt{2}, 1 - \sqrt{2} \)
Explanation :
Given equation \( x^2 - 2x - 1 = 0 \)
By quadratic formula
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{2 \pm \sqrt{4 + 4}}{2} \)
\( x = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \)

Question. The age of a father is 25 years more than his son‘s age. The product of their ages is 84 in years. What will be son‘s age in year, after 10 years?
(a) 3
(b) 28
(c) 13
(d) 18
Answer: (c) 13
Explanation :
Let age of son be \( x \) years
So age of father be \( (x + 25) \) years
\( x(x + 25) = 84 \)
\( x^2 + 25x - 84 = 0 \)
\( x^2 + 28x - 3x - 84 = 0 \)
\( (x + 28)(x - 3) = 0 \)
\( x = - 28 \) and \( x = 3 \)
Son age after 10 years \( (3 + 10) = 13 \) years

Very Short Answer Type Questions

Question. Find the marks obtained by Neha in two subjects, if their average is 75 and their product is 5600.
Answer: Let the marks obtained by Neha in the two subjects be \( x \) and \( y \).
Hence, \( xy = 5600 \) and \( \frac{x + y}{2} = 75 \)
Thus, \( x + y = 150 \)
or \( x = 150 - y \)
and \( xy = 5600 \)
\( \Rightarrow (150 - y)y = 5600 \)
\( \Rightarrow 150y - y^2 = 5600 \)
\( \Rightarrow y^2 - 150y + 5600 = 0 \)
\( \Rightarrow y^2 - 80y - 70y + 5600 = 0 \)
\( \Rightarrow y( y - 80) - 70( y - 80) = 0 \)
\( \Rightarrow ( y - 80) ( y - 70) = 0 \)
\( \Rightarrow y = 70 \) or 80
and \( x = 80 \) or 70

Question. The perimeter of an isosceles triangle is 65 cm and the unequal side is thrice as large as each of the equal sides. Find the lengths of the sides.
Answer: Given, Perimeter of an isosceles triangle = 65 cm
Let the sides of the triangle be \( a \) cm, \( a \) cm and \( b \) cm.
Then \( b = 3a \)
Also \( a + b + a = 65 \)
\( \Rightarrow a + 3a + a = 65 \)
\( \Rightarrow 5a = 65 \)
\( \Rightarrow a = 13 \) cm
and \( b = 39 \) cm

Question. Find the nature of roots of the quadratic equation \( 2x^2 - 4x + 3 = 0 \).*
Answer: Given, \( 2x^2 - 4x + 3 = 0 \)
Comparing it with quadratic equation \( ax^2 + bx + c = 0 \)
Here, \( a = 2, b = - 4 \) and \( c = 3 \)
\( \therefore D = b^2 - 4ac = (- 4)^2 - 4 \times (2) (3) = 16 - 24 = - 8 < 0 \)
Hence, \( D < 0 \) this shows that roots will be imaginary.

Question. Is the roots of the equation \( x^2 - 3x - 9 = 0 \) are real and distinct?
Answer: Given, \( x^2 - 3x - 9 = 0 \)
Now, \( D = b^2 - 4ac \)
\( = (- 3)^2 - 4(1) (- 9) \)
\( = 9 + 36 \)
\( = 45 > 0 \)
Thus, the roots are real and distinct.

Question. For what values of \( k \), the roots of the equation \( x^2 + 4x + k = 0 \) are real ?*
Answer: The given equation is \( x^2 + 4x + k = 0 \)
on comparing the given equation with \( ax^2 + bx + c = 0 \), we get
\( a = 1, b = 4 \) and \( c = k \)
For real roots, \( D \geq 0 \)
or \( b^2 - 4ac \geq 0 \)
\( 16 - 4k \geq 0 \)
or \( k \leq 4 \)
\( \therefore \) For \( k \leq 4 \), equation \( x^2 + 4x + k \) will have real roots.

Question. The solution of a quadratic equation is as follows :
\( x = \frac{8 \pm \sqrt{(- 8)^2 - 4(3)(2)}}{2(3)} \)
Then find the quadratic equation.
Answer: Given, \( x = \frac{8 \pm \sqrt{(- 8)^2 - 4(3)(2)}}{2(3)} \)
Thus \( a = 3, b = - 8 \) and \( c = 2 \)
The equation is given by \( ax^2 + bx + c = 0 \)
Hence \( 3x^2 - 8x + 2 = 0 \)

Question. Solve : \( \sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0 \)
Answer: Given, \( \sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0 \)
\( \Rightarrow x = \frac{2\sqrt{2} \pm \sqrt{(- 2\sqrt{2})^2 - 4(\sqrt{3})(- 2\sqrt{3})}}{2(\sqrt{3})} \)
[ applying the formula : \( x = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a} \) ]
\( \Rightarrow x = \frac{2\sqrt{2} \pm \sqrt{8 + 24}}{2\sqrt{3}} = \frac{2\sqrt{2} \pm \sqrt{32}}{2\sqrt{3}} \)
\( = \frac{\sqrt{2} \pm 2\sqrt{2}}{\sqrt{3}} = \sqrt{6} \) and \( - \sqrt{\frac{2}{3}} \).

Question. Solve : \( abx^2 + (b^2 - ac)x - bc = 0 \).
Answer: Given, \( abx^2 + (b^2 - ac)x - bc = 0 \)
\( \Rightarrow abx^2 + b^2x - acx - bc = 0 \)
\( \Rightarrow bx(ax + b) - c(ax + b) = 0 \)
\( \Rightarrow (ax + b) (bx - c) = 0 \)
\( \Rightarrow x = -\frac{b}{a} \) and \( \frac{c}{b} \).

Question. Solve for \( x \) : \( \sqrt{6x + 7} - (2x - 7) = 0 \)
Answer: We have, \( \sqrt{6x + 7} - (2x - 7) = 0 \)
\( \sqrt{6x + 7} = (2x - 7) \)
Squaring both sides
\( (\sqrt{6x + 7})^2 = (2x - 7)^2 \)
\( \Rightarrow 6x + 7 = 4x^2 + 49 - 28x \)
\( \Rightarrow 4x^2 + 42 - 34x = 0 \)
\( \Rightarrow 2x^2 - 17x + 21 = 0 \)
\( \Rightarrow 2x^2 - 14x - 3x + 21 = 0 \)
\( \Rightarrow 2x (x - 7) - 3 (x - 7) = 0 \)
\( \Rightarrow (2x - 3) (x - 7) = 0 \)
\( \Rightarrow x = \frac{3}{2} \) or 7
\( \therefore x = 7 \) (as \( x = 3/2 \) doesn’t satisfy the given equation)

Question. Solve : \( 12abx^2 - (9a^2 - 8b^2)x - 6ab = 0 \).
Answer: Given, \( 12abx^2 - (9a^2 - 8b^2)x - 6ab = 0 \)
\( \Rightarrow 12abx^2 - 9a^2x + 8b^2x - 6ab = 0 \)
\( \Rightarrow 3ax(4bx - 3a) + 2b(4bx - 3a) = 0 \)
\( \Rightarrow (3ax + 2b) (4bx - 3a) = 0 \)
\( \Rightarrow x = -\frac{2b}{3a} \) and \( \frac{3a}{4b} \).

Question. Solve : \( p^2x^2 + ( p^2 - q^2)x - q^2 = 0 \).
Answer: Given, \( p^2x^2 + ( p^2 - q^2)x - q^2 = 0 \)
\( \Rightarrow p^2x^2 + p^2x - q^2x - q^2 = 0 \)
\( \Rightarrow p^2x(x + 1) - q^2(x + 1) = 0 \)
\( \Rightarrow ( p^2x - q^2)(x + 1) = 0 \)
\( \Rightarrow x = \left(\frac{q}{p}\right)^2 \) and – 1.

Question. If the roots of the equation \( (a - b)x^2 + (b - c)x + (c - a) = 0 \) are equal, prove that \( b + c = 2a \).
Answer: Given, \( (a - b)x^2 + (b - c)x + (c - a) = 0 \)
Comparing with \( Ax^2 + Bx + C = 0 \), we get
\( A = a - b, B = b - c \) and \( C = c - a \)
Since, the roots are equal
\( \therefore D = 0 \)
\( \Rightarrow B^2 - 4AC = 0 \)
\( \Rightarrow (b - c)^2 - 4(a - b)(c - a) = 0 \)
\( \Rightarrow b^2 + c^2 - 2bc - 4(ac - bc - a^2 + ab) = 0 \)
\( \Rightarrow b^2 + c^2 - 2bc - 4ac + 4bc + 4a^2 - 4ab = 0 \)
\( \Rightarrow b^2 + c^2 + 4a^2 + 2bc - 4ac - 4ab = 0 \)
\( \Rightarrow (b + c - 2a)^2 = 0 \)
\( \Rightarrow b + c - 2a = 0 \)
\( \Rightarrow b + c = 2a \).
Hence Proved.

Question. Find the value of \( k \) for which the equation \( x^2 + k(2x + k - 1) + 2 = 0 \) has real and equal roots.
Answer: Given equation is,
\( x^2 + k(2x + k - 1) + 2 = 0 \)
\( \Rightarrow x^2 + 2kx + k(k - 1) + 2 = 0 \)
Here \( a = 1, b = 2k \) and \( c = k(k - 1) + 2 \)
For real and equal roots
\( b^2 - 4ac = 0 \)
\( \Rightarrow (2k)^2 - 4.1.[(k (k - 1) + 2)] = 0 \)
\( \Rightarrow 4k^2 - 4 (k^2 - k + 2) = 0 \)
\( \Rightarrow 4k^2 - 4k^2 + 4k - 8 = 0 \)
\( \Rightarrow 4k = 8 \)
\( \Rightarrow k = \frac{8}{4} = 2 \) Ans.

Question. A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left. When he increased the square by one row and one column, he was short of 25 students. Find the number of students.
Answer: Let the number of rows be \( x \) and the number of students in each row also be \( x \).
Thus \( x^2 + 24 = (x + 1)^2 - 25 \)
\( \Rightarrow (x + 1)^2 - x^2 = 24 + 25 \)
\( \Rightarrow (x + 1 - x)(x + 1 + x) = 49 \)
\( \Rightarrow 2x + 1 = 49 \)
\( \Rightarrow 2x = 48 \)
\( \Rightarrow x = 24 \)
Thus, the total number of students = \( x^2 + 24 \)
\( = (24)^2 + 24 = 24(24 + 1) \)
\( = 24 \times 25 = 600 \).

Question. Solve : \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \).
Answer: Given,
\( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \)
\( \Rightarrow 4x^2 - 4a^2x + (a^2 - b^2)(a^2 + b^2) = 0 \)
\( \Rightarrow 4x^2 - 2(a^2 - b^2)x - 2(a^2 + b^2)x + (a^2 - b^2)(a^2 + b^2) = 0 \)
\( \Rightarrow 2x[2x - (a^2 - b^2)] - (a^2 + b^2) [2x - (a^2 - b^2)] = 0 \)
\( \Rightarrow [2x - (a^2 - b^2)][2x - (a^2 + b^2)] = 0 \)
\( \Rightarrow x = \frac{a^2 - b^2}{2} \) and \( \frac{a^2 + b^2}{2} \).

Question. Solve : \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \).
Answer: Given,
\( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
\( \Rightarrow \sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0 \)
\( \Rightarrow x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0 \)
\( \Rightarrow (\sqrt{2}x + 5) (x + \sqrt{2}) = 0 \)
\( \Rightarrow x = - \frac{5}{\sqrt{2}} \) and \( - \sqrt{2} \).

Question. Find the value of \( p \), for which one root of the quadratic equation \( px^2 - 14x + 8 = 0 \) is 6 times the other.
Answer: Given equation is,
\( px^2 - 14x + 8 = 0 \)
Let one root = \( \alpha \)
Then other root = \( 6\alpha \)
Sum of roots = \( -\frac{b}{a} \)
\( \Rightarrow \alpha + 6\alpha = \frac{-(-14)}{p} \)
\( \Rightarrow 7\alpha = \frac{14}{p} \)
\( \Rightarrow \alpha = \frac{14}{p \times 7} \)
or \( \alpha = \frac{2}{p} \) ...(i)
Product of roots = \( \frac{c}{a} \)
\( \Rightarrow (\alpha) (6\alpha) = \frac{8}{p} \)
\( \Rightarrow 6\alpha^2 = \frac{8}{p} \) ...(ii)
Putting value from equation (i) to equation (ii), we get
\( 6 \left(\frac{2}{p}\right)^2 = \frac{8}{p} \)
\( \Rightarrow 6 \times \frac{4}{p^2} = \frac{8}{p} \)
\( \Rightarrow 24p = 8p^2 \)
\( \Rightarrow 8p^2 - 24p = 0 \)
\( \Rightarrow 8p ( p - 3) = 0 \)
Either \( 8p = 0 \Rightarrow p = 0 \)
or \( p - 3 = 0 \Rightarrow p = 3 \)
\( \therefore p = 3 \) (as \( p = 0 \) doesn’t satisfy the given equation)

Question. If – 5 is a root of the equation \( 2x^2 + px - 15 = 0 \) and the quadratic equation \( p(x^2 + x) + k = 0 \) has equal roots, find the value of \( k \).
Answer: As – 5 is a root of the equation \( 2x^2 + px - 15 = 0 \),
\( \therefore 2(- 5)^2 + p(- 5) - 15 = 0 \)
\( \Rightarrow 2(25) - 5p - 15 = 0 \)
\( \Rightarrow 5p = 50 - 15 \)
\( \Rightarrow 5p = 35 \)
\( \Rightarrow p = 7 \)
Now substituting \( p = 7 \) in \( p(x^2 + x) + k = 0 \), we have
\( 7(x^2 + x) + k = 0 \)
\( \Rightarrow 7x^2 + 7x + k = 0 \)
Since, the above equation has equal roots,
\( D = (7)^2 - 4(7)(k) = 0 \)
\( \Rightarrow 49 - 28k = 0 \)
\( \Rightarrow 28k = 49 \)
\( \Rightarrow 4k = 7 \)
\( k = \frac{7}{4} \)
Thus, the equation will have real roots, if \( k = \frac{7}{4} \).

Question. Find the value of \( k \) for which the roots of the equation \( 3x^2 - 10x + k = 0 \) are reciprocal of each other.*
Answer: The given equation is \( 3x^2 - 10x + k = 0 \)
On comparing it with \( ax^2 + bx + c = 0 \), we get
\( a = 3, b = - 10, c = k \)
Let the roots of the equation are \( \alpha \) and \( \frac{1}{\alpha} \)
Product of the roots = \( \frac{c}{a} \)
\( \therefore \alpha \cdot \frac{1}{\alpha} = \frac{k}{3} \)
or \( k = 3 \)

Question. Find the value of \( k \) for which the equation \( x^2 - 4kx + k = 0 \) has real and equal roots.
Answer: Given,
\( x^2 - 4kx + k = 0 \)
Since, the equation has real and equal roots,
\( D = (- 4k)^2 - 4(1)(k) = 0 \)
\( \Rightarrow 16k^2 - 4k = 0 \)
\( \Rightarrow 4k(4k - 1) = 0 \)
\( \Rightarrow k = 0, \frac{1}{4} \)
Thus, the equation will have real roots, if \( k = 0, \frac{1}{4} \).

Question. If the quadratic equation \( px^2 - 2\sqrt{5}px + 15 = 0 \), has two equal roots then find the value of \( p \).
Answer: The given quadratic equation is,
\( px^2 - 2\sqrt{5}px + 15 = 0 \)
This equation is of the form
\( ax^2 + bx + c = 0 \)
where, \( a = p, b = - 2\sqrt{5}p, c = 15 \)
We know , \( D = b^2 - 4ac \)
\( = (- 2\sqrt{5}p)^2 - 4 \times p \times 15 \)
\( = 20p^2 - 60p \)
\( = 20p( p - 3) \)
For real and equal roots, we must have
\( D = 0 \Rightarrow 20p( p - 3) = 0 \)
\( \Rightarrow p = 0, p = 3 \)
\( p = 0 \), is not possible as whole equation will be zero.
Hence, 3 is the required value of \( p \).

Question. Find two consecutive numbers whose squares have the sum 85.
Answer: Let the two numbers be \( x \) and \( (x + 1) \).
Thus, \( x^2 + (x + 1)^2 = 85 \)
\( \Rightarrow x^2 + x^2 + 2x + 1 = 85 \)
\( \Rightarrow 2x^2 + 2x - 84 = 0 \)
\( \Rightarrow x^2 + x - 42 = 0 \)
\( \Rightarrow x^2 + 7x - 6x - 42 = 0 \)
\( \Rightarrow x(x + 7) - 6(x + 7) = 0 \)
\( \Rightarrow (x + 7)(x - 6) = 0 \)
\( \Rightarrow x = 6, - 7 \)
Thus, the two numbers are either 6 and 7 or – 7 and – 6.

Question. A peacock is sitting on the top of a pillar which is 9 m high. From a point, 27 m away from the bottom of a pillar, a snake is coming to its hole at the base of the pillar. Seeing the snake, the peacock pounces on it. If their speeds are equal, at what distance from the hole is the snake caught ?
Answer: Let the distance, where the snake is caught from the base of the pillar be \( x \) m. So the distance travelled by the snake before being caught is \( (27 - x) \) m.
As the speed of the peacock and the snake is the same, so the distance travelled by the two is also same. Thus, the peacock travels \( (27 - x) \) m from the top of the tower to catch the snake. Hence \( (27 - x) \) is the hypotenuse. Applying Pythagoras theorem,
\( (9)^2 + x^2 = (27 - x)^2 \)
\( \Rightarrow x^2 + 81 = 729 + x^2 - 54x \)
\( \Rightarrow 54x = 729 - 81 \)
\( \Rightarrow 54x = 648 \)
\( \Rightarrow x = 12 \) m
Hence, the snake is caught 12 m from the pillar.

Question. Show that \( x = - 2 \) is a solution of \( 3x^2 + 13x + 14 = 0 \).
Answer: The given equation is,
\( 3x^2 + 13x + 14 = 0 \)
Substituting \( x = - 2 \), we have
L.H.S. \( = 3(- 2)^2 + 13(- 2) + 14 \)
\( = 12 - 26 + 14 \)
\( = 0 = \) R.H.S.
Thus, \( x = - 2 \) is a solution of \( 3x^2 + 13x + 14 = 0 \).
Hence Proved.

Question. Solve the following quadratic equation for \( x \) :
\( 9x^2 - 6b^2x - (a^4 - b^4) = 0 \)
Answer: \( 9x^2 - 6b^2x - (a^4 - b^4) = 0 \)
\( \Rightarrow x = \frac{6b^2 \pm \sqrt{36b^4 + 4 \times 9 \times (a^4 - b^4)}}{2 \times 9} \)
\( \Rightarrow x = \frac{6b^2 \pm \sqrt{36b^4 + 36a^4 - 36b^4}}{2 \times 9} \)
\( \Rightarrow x = \frac{6b^2 \pm \sqrt{36a^4}}{2 \times 9} \)
\( \Rightarrow x = \frac{6b^2 \pm 6a^2}{2 \times 3 \times 3} \)
\( \Rightarrow x = \frac{b^2 \pm a^2}{3} \)
\( \Rightarrow x = \frac{b^2 + a^2}{3}, \frac{b^2 - a^2}{3} \)

Question. Find the value of \( p \) for which the quadratic equation \( (p + 1)x^2 - 6(p + 1)x + 3(p + 9) = 0, p \neq - 1 \) has equal roots. Hence, find the roots of the equation.
Answer: \( (p + 1)x^2 - 6(p + 1)x + 3(p + 9) = 0 \)
\( p \neq - 1 \)
\( \therefore \) Above equation has equal roots.
So, discriminant,
\( D = 0 \)
\( \Rightarrow \{- 6(p + 1)\}^2 - 4 \times (p + 1) \cdot 3(p + 9) = 0 \)
\( \Rightarrow 36(p^2 + 2p + 1) - 12(p^2 + 10p + 9) = 0 \)
\( \Rightarrow 24p^2 - 48p - 72 = 0 \)
\( \Rightarrow p^2 - 2p - 3 = 0 \)
\( \Rightarrow (p - 3)(p + 1) = 0 \)
\( p = 3 \) (as \( p \neq - 1 \))
The given quadratic equation becomes,
\( 4x^2 - 24x + 36 = 0 \)
\( \Rightarrow x^2 - 6x + 9 = 0 \)
\( \Rightarrow (x - 3)^2 = 0 \)
\( \Rightarrow x = 3, 3 \)
Hence, roots of given quadratic equation are 3 and 3.

Question. Solve for \( x \) :
\( \frac{2}{x + 1} + \frac{3}{2(x - 2)} = \frac{23}{5x} \), \( x \neq 0, - 1, 2 \)
Answer: \( \frac{2}{x + 1} + \frac{3}{2(x - 2)} = \frac{23}{5x} \)
\( \Rightarrow \frac{4(x - 2) + 3(x + 1)}{2(x + 1)(x - 2)} = \frac{23}{5x} \)
\( \Rightarrow \frac{7x - 5}{2x^2 - 2x - 4} = \frac{23}{5x} \)
\( \Rightarrow 35x^2 - 25x = 46x^2 - 46x - 92 \)
\( \Rightarrow 11x^2 - 21x - 92 = 0 \)
\( \Rightarrow x = \frac{21 \pm \sqrt{(- 21)^2 - 4 \times 11 \times (- 92)}}{2 \times 11} \)
\( \Rightarrow x = \frac{21 \pm \sqrt{441 + 4048}}{2 \times 11} \)
\( \Rightarrow x = \frac{21 \pm \sqrt{4489}}{22} \)
\( \Rightarrow x = \frac{21 \pm 67}{22} \)
\( \Rightarrow x = \frac{21 + 67}{22}, \frac{21 - 67}{22} \)
\( \Rightarrow x = 4, -\frac{23}{11} \)
Hence, \( x = 4 \) or \( -\frac{23}{11} \)

Question. The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is 29/20. Find the original fraction.
Answer: Let denominator of fraction \( = x \)
Then, numerator \( = x - 3 \)
\( \therefore \) Fraction \( = \frac{x - 3}{x} \)
New fraction \( = \frac{x - 3 + 2}{x + 2} = \frac{x - 1}{x + 2} \)
According to question,
\( \Rightarrow \frac{x - 1}{x + 2} + \frac{x - 3}{x} = \frac{29}{20} \)
\( \Rightarrow \frac{x(x - 1) + (x + 2)(x - 3)}{(x + 2)(x)} = \frac{29}{20} \)
\( \Rightarrow \frac{x^2 - x + x^2 - 3x + 2x - 6}{x^2 + 2x} = \frac{29}{20} \)
\( \Rightarrow \frac{2x^2 - 2x - 6}{x^2 + 2x} = \frac{29}{20} \)
\( \Rightarrow 40x^2 - 40x - 120 = 29x^2 + 58x \)
\( \Rightarrow 40x^2 - 40x - 120 - 29x^2 - 58x = 0 \)
\( \Rightarrow 11x^2 - 98x - 120 = 0 \)
\( \Rightarrow 11x^2 - 110x + 12x - 120 = 0 \)
\( \Rightarrow 11x(x - 10) + 12(x - 10) = 0 \)
\( \Rightarrow (x - 10)(11x + 12) = 0 \)
\( \Rightarrow x - 10 = 0 \)
\( \Rightarrow x = 10 \)
\( \Rightarrow 11x + 12 = 0 \)
\( \Rightarrow x = \frac{- 12}{11} \) (rejected)
Hence, fraction is \( \frac{x - 3}{x} = \frac{10 - 3}{10} = \frac{7}{10} \)

Question. The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.
Answer: Let shorter side \( = x \) m
\( \therefore \) Diagonal \( = (x + 16) \) m
and Longer side \( = (x + 14) \) m
Now, by Pythagoras theorem,
\( (x + 16)^2 = (x + 14)^2 + x^2 \)
\( x^2 + 32x + 256 = x^2 + 28x + 196 + x^2 \)
\( \Rightarrow x^2 - 4x - 60 = 0 \)
\( \Rightarrow (x - 10)(x + 6) = 0 \)
\( \Rightarrow x = 10 \) or \( x = - 6 \) (Rejected)
\( \therefore \) Shorter side \( = 10 \) m
Diagonal \( = 26 \) m
and Longer side \( = 24 \) m

Question. Find the value of \( p \) for which the quadratic equation \( (2p + 1)x^2 - (7p + 2)x + (7p - 3) = 0 \) has equal roots. Also find these roots.
Answer: \( (2p + 1)x^2 - (7p + 2)x + (7p - 3) = 0 \)...(i)
\( \therefore \) Above quadratic equation has equal roots,
\( \therefore D = 0 \)
\( \Rightarrow \{- (7p + 2)\}^2 - 4(2p + 1)(7p - 3) = 0 \)
\( \Rightarrow 49p^2 + 4 + 28p - 4(14p^2 + p - 3) = 0 \)
\( \Rightarrow 49p^2 + 4 + 28p - 56p^2 - 4p + 12 = 0 \)
\( \Rightarrow - 7p^2 + 29p + 16 = 0 \)
\( \Rightarrow 7p^2 - 24p - 16 = 0 \)
\( \Rightarrow (p - 4)(7p + 4) = 0 \)
\( \Rightarrow p = 4 \) or \( p = -\frac{4}{7} \)
When \( p = -\frac{4}{7} \), the given quadratic equation (i) becomes
\( \left(-\frac{8}{7} + 1\right)x^2 - \left(-\frac{28}{7} + 2\right)x + \left(-\frac{28}{7} - 3\right) = 0 \)
\( \Rightarrow - x^2 + 14x - 49 = 0 \)
\( \Rightarrow x^2 - 14x + 49 = 0 \)
\( \Rightarrow (x - 7)^2 = 0 \)
\( \Rightarrow x = 7, 7 \)
When \( p = 4 \), then the quadratic equation (i) becomes
\( (8 + 1)x^2 - (28 + 2)x + (28 - 3) = 0 \)
\( \Rightarrow 9x^2 - 30x + 25 = 0 \)
\( \Rightarrow (3x - 5)^2 = 0 \)
\( \Rightarrow x = \frac{5}{3}, \frac{5}{3} \)
Hence, roots are 7, 7 or \( \frac{5}{3}, \frac{5}{3} \).

Question. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Answer: Let two consecutive odd numbers be \( x \) and \( x + 2 \).
\( \therefore x^2 + (x + 2)^2 = 394 \)
\( \Rightarrow x^2 + x^2 + 4x + 4 = 394 \)
\( \Rightarrow 2x^2 + 4x - 390 = 0 \)
\( \Rightarrow x^2 + 2x - 195 = 0 \)
\( \Rightarrow x^2 + 15x - 13x - 195 = 0 \)
\( \Rightarrow x(x + 15) - 13(x + 15) = 0 \)
\( \Rightarrow (x - 13)(x + 15) = 0 \)
\( \Rightarrow x - 13 = 0 \) or \( x + 15 = 0 \)
\( \Rightarrow x = 13 \) or \( x = - 15 \)
\( \therefore \) Numbers are 13 and \( 13 + 2 = 15 \), i.e., 13 and 15 or – 15 and \( - 15 + 2 = - 13 \) i.e., – 15 and – 13.

Question. Sum of the areas of two squares is \( 400 \text{ cm}^2 \). If the difference of their perimeters is 16 cm, find the sides of the two squares.
Answer: Let, side of one square be \( x \)
Side of another square be \( y \)
Sum of areas of two squares \( = x^2 + y^2 \)
\( \therefore x^2 + y^2 = 400 \)...(i)
Difference of their perimeters \( = 4x - 4y \)
\( \therefore 4x - 4y = 16 \)
\( \Rightarrow x - y = 4 \)
\( \Rightarrow y = x - 4 \)[Put in (i)]
\( \therefore \) Equation (i) becomes :
\( x^2 + (x - 4)^2 = 400 \)
\( \Rightarrow 2x^2 - 8x + 16 = 400 \)
\( \Rightarrow 2x^2 - 8x - 384 = 0 \)
\( \Rightarrow x^2 - 4x - 192 = 0 \)
\( (x - 16)(x + 12) = 0 \)
\( \Rightarrow x = 16, x = - 12 \) (not possible as side cannot be negative)
\( \therefore y = 16 - 4 = 12 \)
Sides of squares are 12 cm and 16 cm.