CBSE Class 10 Mathematics Quadratic Equations VBQs Set B

Multiple Choice Question

Question. The value(s) of k for which the quadratic equation \(2x^2 + kx + 2 = 0\) has equal roots, is:
(a) 4
(b) ± 4
(c) – 4
(d) 0
Answer: (b)
Explanation: The equation \(2x^2 + kx + 2 = 0\) has equal roots, when \(D = (k)^2 – 4(2)(2) = 0\) (\(\because D = b^2 – 4ac\))
\(\Rightarrow k^2 = 16\) or \(k = \pm 4\)

Question. The quadratic equations \(x^2 – 4x + k = 0\) has distinct real roots if
(a) \(k = 4\)
(b) \(k > 4\)
(c) \(k = 16\)
(d) \(k < 4\)
Answer: (d)
Explanation: Equation \(x^2 – 4x + k = 0\) will have distinct real roots, if \(D = (–4)^2 – 4(k)(1) > 0\) [\(\because D = b^2 – 4ac\)]
i.e., \(16 – 4k > 0\) or \(k < 4\)

Question. The equations \(x^2 – 8x + k = 0\) has real and distinct roots if:
(a) \(k = 16\)
(b) \(k > 16\)
(c) \(k = 8\)
(d) \(k < 16\)
Answer: (d)
Explanation: Here, \(a = 1, b = – 8, c = k\), Then, \(D = (– 8)^2 – 4k\) [\(\because D = b^2 – 4ac\)]
For the equation to have real and distinct roots, \(D\) must be greater than \(0\), i.e., \(64 – 4k > 0\) or \(k < 16\)

Question. Which of the following is a quadratic equation?
(a) \(x^2 + 2x + 1 = (4 – x)^2 + 3\)
(b) \(–2x^2 = (5 – x) \left(2x - \frac{2}{5}\right)\)
(c) \((k + 1)x^2 + \frac{3}{2}x = 7\), where \(k = –1\)
(d) \(x^3 – x^2 = (x – 1)^3\)
Answer: (d)
Explanation: Given equation is
(a) \(x^2 + 2x + 1 = (4 – x)^2 + 3\)
\(\Rightarrow x^2 + 2x + 1 = 16 + x^2 – 8x + 3\)
\(\Rightarrow 10x = 18 \Rightarrow 10x – 18 = 0\)
This equation is not of the form \(ax^2 + bx + c, a \neq 0\). Thus, it is not a quadratic equation.
(b) \(–2x^2 = (5 – x) \left(2x - \frac{2}{5}\right)\)
\(\Rightarrow –2x^2 = 10x – 2x^2 – 2 + \frac{2x}{5}\)
\(\Rightarrow 50x + 2x – 10 = 0 \Rightarrow 52x – 10 = 0\)
This is also not a quadratic equation as it is also not of the form \(ax^2 + bx + c = 0, a \neq 0\).
(c) \((k + 1)x^2 + \frac{3}{2}x = 7\), where \(k = -1\)
\(\Rightarrow (–1 + 1)x^2 + \frac{3x}{2} = 7 \Rightarrow \frac{3x}{2} = 7 \Rightarrow 3x – 14 = 0\)
This is also not a quadratic equation as it is also not of the form \(ax^2 + bx + c = 0, a \neq 0\).
(d) \(x^3 – x^2 = (x – 1)^3\)
\(x^3 – x^2 = x^3 – 3x^2(1) + 3x(1)^2 – (1)^3\) [\(\because (a – b)^3 = a^3 – 3a^2b + 3b^2a – b^3\)]
\(\Rightarrow x^3 – x^2 = x^3 – 3x^2 + 3x – 1\)
\(\Rightarrow –x^2 + 3x^2 – 3x + 1 = 0 \Rightarrow 2x^2 – 3x + 1 = 0\)
This represents a quadratic equation because it is of the form \(ax^2 + bx + c = 0, a \neq 0\).

Question. Which of the following is not a quadratic equation?
(a) \(2(x – 1)^2 = 4x^2 – 2x + 1\)
(b) \(2x – x^2 = x^2 + 5\)
(c) \((\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 - 5x\)
(d) \((x^2 + 2x)^2 = x^4 + 3 + 4x^3\)
Answer: (c)
Explanation: It is given that
(a) \(2(x – 1)^2 = 4x^2 – 2x + 1\)
\(\Rightarrow 2(x^2 + 1 – 2x) = 4x^2 – 2x + 1\)
\(\Rightarrow 2x^2 + 2 – 4x = 4x^2 – 2x + 1\)
\(\Rightarrow 2x^2 + 2x – 1 = 0\)
This represents a quadratic equation as it is of the form \(ax^2 + bx + c = 0, a \neq 0\).
(b) \(2x – x^2 = x^2 + 5\)
\(\Rightarrow x^2 + 5 + x^2 – 2x = 0 \Rightarrow 2x^2 – 2x + 5 = 0\)
This also represents a quadratic equation.
(c) \((\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 – 5x\)
\(\Rightarrow 2x^2 + 3 + 2\sqrt{6}x + x^2 = 3x^2 – 5x\)
\(\Rightarrow 3x^2 + 3 + 2\sqrt{6}x – 3x^2 + 5x = 0 \Rightarrow (5 + 2\sqrt{6})x + 3 = 0\)
This does not represent a quadratic equation as it is not of the form \(ax^2 + bx + c = 0, a \neq 0\).
(d) \((x^2 + 2x)^2 = x^4 + 3 + 4x^3\)
\(x^4 + 4x^2 + 4x^3 = x^4 + 3 + 4x^3 \Rightarrow 4x^2 – 3 = 0\)
This represents a quadratic equation.

Definition

• An equation which is of the form \(ax^2 + bx + c = 0\), \(a \neq 0\) is called a quadratic equation.

Question. For what value of k, \(kx^2 + 8x + 2 = 0\) has real roots
(a) \(k < 8\)
(b) \(k > 8\)
(c) \(k = 8\)
(d) none of these
Answer: (a)
Explanation: \(kx^2 + 8x + 2 = 0\). Here, \(a = k, b = 8, c = 2\). For real roots \(b^2 – 4ac > 0\)
\(\Rightarrow 8^2 – 4k \times 2 > 0 \Rightarrow 64 – 8k > 0 \Rightarrow 8 – k > 0 \Rightarrow k < 8\)

Question. Which of the following equations has 2 as a root?
(a) \(x^2 – 4x + 5 = 0\)
(b) \(x^2 + 3x – 12 = 0\)
(c) \(2x^2 – 7x + 6 = 0\)
(d) \(3x^2 – 6x – 2 = 0\)
Answer: (c)
Explanation:
(a) Putting the value of \(x = 2\) in \(x^2 – 4x + 5 = 0\), we get \((2)^2 – 4(2) + 5 = 0 \Rightarrow 4 – 8 + 5 = 1 \neq 0\). So, \(x = 2\) is not a root of \(x^2 – 4x + 5 = 0\).
(b) Putting the value of \(x = 2\) in \(x^2 + 3x – 12 = 0\), we get \((2)^2 + 3(2) – 12 = 0 \Rightarrow 4 + 6 – 12 = –2 \neq 0\). So, \(x = 2\) is not a root of \(x^2 + 3x – 12 = 0\).
(c) Putting the value of \(x = 2\) in \(2x^2 – 7x + 6 = 0\), we get \(2(2)^2 – 7(2) + 6 = 0 \Rightarrow 8 – 14 + 6 = 0\). So, \(x = 2\) is the root of the equation \(2x^2 – 7x + 6 = 0\).
(d) Putting the value of \(x = 2\) in \(3x^2 – 6x – 2 = 0\), we get \(3(2)^2 – 6(2) – 2 = –2 \neq 0\). So, \(x = 2\) is not the root of the equation \(3x^2 – 6x – 2 = 0\).

Trick Applied

• A real number \(\alpha\) is said to be the root of the quadratic equation \(ax^2 + bx + c = 0\) if \(a\alpha^2 + b\alpha + c = 0\).

Question. The positive root of \(\sqrt{3x^2 + 6} = 9\) is
(a) 2
(b) 1
(c) 4
(d) 3
Answer: (b)
Explanation: \(\sqrt{3x^2 + 6} = 9\). Squaring both sides \(3x^2 + 6 = 9 \Rightarrow 3x^2 = 3 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1\). Positive root is 1.

Question. Which of the following equations has the sum of its roots as 3?
(a) \(2x^2 – 3x + 6 = 0\)
(b) \(–x^2 + 3x – 3 = 0\)
(c) \(\sqrt{2}x^2 – \frac{3}{\sqrt{2}}x + 1 = 0\)
(d) \(3x^2 – 3x + 3 = 0\)
Answer: (b)
Explanation: We know that sum of the roots = \(-\frac{b}{a}\). On comparing the given equations with \(ax^2 + bx + c = 0\):
(a) \(2x^2 – 3x + 6 = 0 \Rightarrow a = 2, b = –3, c = 6\). Sum of its roots = \(-\frac{b}{a} = -\frac{(-3)}{2} = \frac{3}{2}\).
(b) \(–x^2 + 3x – 3 = 0 \Rightarrow a = –1, b = 3, c = –3\). Sum of its roots = \(-\frac{b}{a} = -\frac{3}{-1} = 3\).
(c) \(\sqrt{2}x^2 – \frac{3}{\sqrt{2}}x + 1 = 0 \Rightarrow a = \sqrt{2}, b = -\frac{3}{\sqrt{2}}, c = 1\). Sum of its roots = \(-\frac{b}{a} = -\frac{-3/\sqrt{2}}{\sqrt{2}} = \frac{3}{2}\).
(d) \(3x^2 – 3x + 3 = 0 \Rightarrow a = 3, b = –3, c = 3\). Sum of its roots = \(-\frac{b}{a} = -\frac{(-3)}{3} = 1\).

Question. Is \(x^3 – 4x^2 – x + 1 = (x – 2)^3\) a quadratic equation?
(a) yes
(b) No
(c) Can’t say
(d) This is a cubic equation
Answer: (a)
Explanation: \(x^3 – 4x^2 – x + 1 = (x – 2)^3 \Rightarrow x^3 – 4x^2 – x + 1 = x^3 – 8 – 6x^2 + 12x \Rightarrow 2x^2 – 13x + 9 = 0\). This is a quadratic equation.

Question. For equal root, \(kx(x – 2) + 6 = 0\), the value of k is
(a) \(k = 0, 6\)
(b) \(k = 6, – 6\)
(c) \(k = 2, 3\)
(d) \(k = 0, 3\)
Answer: (a)
Explanation: We have \(kx(x – 2) + 6 = 0 \Rightarrow kx^2 – 2kx + 6 = 0\). Here \(a = k, b = – 2k, c = 6\). For equal roots, \(b^2 – 4ac = 0\).
\((– 2k)^2 – 4k \times 6 = 0 \Rightarrow 4k^2 – 24k = 0 \Rightarrow 4k (k – 6) = 0 \Rightarrow k = 0, 6\)

Question. Roots of \(-x^2 + \frac{1}{2}x + \frac{1}{2} = 0\) are
(a) \(-\frac{1}{2}, 1\)
(b) \(\frac{1}{2}, 1\)
(c) \(\frac{1}{2}, -1\)
(d) \(\frac{1}{2}, \frac{1}{2}\)
Answer: (a)
Explanation: \(-x^2 + \frac{1}{2}x + \frac{1}{2} = 0 \Rightarrow – 2x^2 + x + 1 = 0 \Rightarrow – 2x^2 + 2x – x + 1 = 0 \Rightarrow – 2x (x – 1) – 1 (x – 1) = 0 \Rightarrow (x – 1) (– 2x – 1) = 0 \Rightarrow x = 1, x = -\frac{1}{2}\)

Question. The quadratic equation \(2x^2 – \sqrt{5}x + 1 = 0\) has:
(a) two distinct real roots
(b) two equal real roots
(c) no real roots
(d) more than 2 real roots
Answer: (c)
Explanation: We know that if \(D = b^2 – 4ac < 0\) for a quadratic equation \(ax^2 + bx + c = 0\), then roots are not real. The given equation is: \(2x^2 – \sqrt{5}x + 1 = 0\). On comparing it with \(ax^2 + bx + c = 0\), we get \(a = 2, b = –\sqrt{5}, c = 1\). Discriminant, \(D = b^2 – 4ac = (– \sqrt{5})^2 – 4(2)(1) = 5 – 8 = –3 \Rightarrow D = –3 < 0\). Since the discriminant is negative, the given equation has no real roots.

Question. Which of the following equations has two distinct real roots?
(a) \(2x^2 – 3\sqrt{2}x + \frac{9}{4} = 0\)
(b) \(x^2 + x – 5 = 0\)
(c) \(x^2 + 3x + 2\sqrt{2} = 0\)
(d) \(5x^2 – 3x + 1 = 0\)
Answer: (b)
Explanation: We know that if \(D = b^2 – 4ac > 0\) for a quadratic equation \(ax^2 + bx + c = 0\), then its roots are real and distinct.
(a) \(2x^2 – 3\sqrt{2}x + \frac{9}{4} = 0 \Rightarrow a = 2, b = –3\sqrt{2}, c = \frac{9}{4}\). \(D = b^2 – 4ac = (–3\sqrt{2})^2 – 4(2)(\frac{9}{4}) = 18 – 18 = 0\). Equal roots.
(b) \(x^2 + x – 5 = 0 \Rightarrow a = 1, b = 1, c = –5\). \(D = b^2 – 4ac = (1)^2 – 4(1)(–5) = 1 + 20 = 21 > 0\). Two distinct real roots.
(c) \(x^2 + 3x + 2\sqrt{2} = 0 \Rightarrow a = 1, b = 3, c = 2\sqrt{2}\). \(D = b^2 – 4ac = (3)^2 – 4(1)(2\sqrt{2}) = 9 – 8\sqrt{2} < 0\). No real roots.
(d) \(5x^2 – 3x + 1 = 0 \Rightarrow a = 5, b = –3, c = 1\). \(D = b^2 – 4ac = (–3)^2 – 4(5)(1) = 9 – 20 = –11 < 0\). No real roots.

Question. \((x^2 + 1)^2 – x^2 = 0\) has:
(a) four real roots
(b) two real roots
(c) no real root
(d) one real root
Answer: (c)
Explanation: The given equation is: \((x^2 + 1)^2 – x^2 = 0 \Rightarrow x^4 + 1 + 2x^2 – x^2 = 0 \Rightarrow x^4 + x^2 + 1 = 0\). Let \(x^2 = y \Rightarrow y^2 + y + 1 = 0\). Comparing with \(ay^2 + by + c = 0\), we get \(a = 1, b = 1, c = 1\). \(D = b^2 – 4ac = (1)^2 – 4(1)(1) = –3 < 0\). As \(D < 0\), the equation has no real roots.

Fill in the Blanks

Question. The quadratic equation \(2x^2 + px + 3 = 0\) has two equal roots if p = .................... .
Answer: \(\pm 2\sqrt{6}\)
Explanation: \(2x^2 + px + 3 = 0\) will have equal roots, when \(p^2 – 4(2)(3) = 0 \Rightarrow p^2 – 24 = 0 \Rightarrow p = \pm \sqrt{24} = \pm 2\sqrt{6}\)

Question. Equation \(ax^2 + bx + c = 0\) represents a quadratic equation if and only if .................... .
Answer: \(a \neq 0\)
Explanation: In case \(a = 0\), the equation reduces to \(bx + c = 0\), which is a linear equation.

Question. Sum of roots of quadratic equation \(x^2 – 4x + 2 = 0\) is ................... of product of roots.
Answer: Twice
Explanation: Sum of roots \(= -\frac{-4}{1} = 4\). Product of roots \(= \frac{2}{1} = 2\). Since \(4 = 2(2)\), sum is twice the product.

Question. The quadratic equation \(2x^2 + x + 4\) has .................... real roots.
Answer: No
Explanation: Here, \(b^2 – 4ac = 1 – 32 = -31\), which is less than 0.

Question. The roots of \(x + \frac{1}{x} = 2\) are ........................ .
Answer: [1, 1]
Explanation: The equation is \(x^2 – 2x + 1 = 0\), or \((x – 1)^2 = 0\). So, roots are 1, 1.

Question. The sum of the roots of the quadratic equation \(2x^2 + 14x + 24 = 0\) is .................... .
Answer: [–7]
Explanation: Sum of roots \(= -\frac{14}{2} = -7\).

Very Short Answer Type Questions

Question. Find the values of ‘k’ for which \(x = 2\) is a solution of the equation \(kx^2 + 2x – 3 = 0\).
Answer: Given equation is, \(kx^2 + 2x – 3 = 0\). If \(x = 2\), then \(k(2)^2 + 2(2) – 3 = 0 \Rightarrow 4k + 1 = 0 \Rightarrow 4k = –1 \Rightarrow k = -\frac{1}{4}\). Hence, the value of \(k\) is \(-\frac{1}{4}\).

Question. Find the value/s of \(k\) for which the quadratic equation \(3x^2 + kx + 3 = 0\) has real and equal roots. 
Answer: Given, quadratic equation is: \(3x^2 + kx + 3 = 0\)
For real and equal roots \(b^2 - 4ac = 0\)
Here, \(a = 3, b = k\) and \(c = 3\)
\(\therefore b^2 - 4ac = (k)^2 - 4 \times 3 \times 3 = 0\)
\(\Rightarrow k^2 = 36\)
\(\Rightarrow k = \pm 6\)
Hence, the value of \(k\) is \(\pm 6\).

Question. For what values of \(k\) does the quadratic equation \(4x^2 - 12x - k = 0\) have no real roots? 
Answer: Given equation is: \(4x^2 - 12x - k = 0\)
For equation to have no real roots, \(D < 0\)
or \(b^2 - 4ac < 0\)
Here, \(a = 4, b = -12, c = -k\)
\((-12)^2 - 4 \times 4 \times (-k) < 0\)
\(144 + 16k < 0\)
\(16k < -144\)
\(k < -9\)
Hence, the value of \(k\) should be less than \(-9\).

Question. Find the nature of the roots of the quadratic equation \(2x^2 - 4x + 3 = 0\). 
Answer: Given: quadratic equation:
\(2x^2 - 4x + 3 = 0\)
Here, \(a = 2, b = -4, c = 3\)
Discriminant, \(D = b^2 - 4ac\)
\(= (-4)^2 - 4 \times 2 \times 3\)
\(= 16 - 24 = -8 < 0\)
\(\therefore D < 0\)
Hence, the given equation does not have real roots.

Question. Is 0.2 a root of the equation \(x^2 - 0.4 = 0\)? Justify. 
Answer: No, because 0.2 does not satisfy the quadratic equation i.e.
\(x^2 - 0.4 = (0.2)^2 - 0.4\)
\(= 0.04 - 0.4\)
\(= -0.36 \neq 0\).

Question. For what values of \(k\), the roots of the equation \(x^2 + 4x + k = 0\) are real?
Answer: Since, the roots of the equation \(x^2 + 4x + k = 0\) are real,
i.e. \(D \geq 0\)
\(b^2 - 4ac \geq 0\)
Here, \(a = 1, b = 4, c = k\)
\(\Rightarrow (4)^2 - 4 \times 1 \times k \geq 0\)
\(\Rightarrow 16 - 4k \geq 0\)
\(\Rightarrow k \leq 4\)
Hence, the value of '\(k\)' is less than or equal to 4.

Question. If \(x = 2\) and \(m = 3\), the equation is \(3x^2 - 2kx + 2m = 0\), find \(k\).
Answer: \(3x^2 - 2kx + 2m = 0\)
\(x = 2\) and \(m = 3\) [Given]
So, \(3(2)^2 - 2k(2) + 2(3) = 0\)
\(\Rightarrow 12 - 4k + 6 = 0\)
\(\Rightarrow -4k + 18 = 0\)
\(\Rightarrow k = \frac{9}{2}\).

Question. If one root of the quadratic equation \(6x^2 - x - k = 0\) is \(\frac{2}{3}\), then find the value of '\(k\)'. 
Answer: Given: quadratic equation: \(6x^2 - x - k = 0\).
Its one of its root: \(\frac{2}{3}\)
If \(x = \frac{2}{3}\) is root of the given equation, then it will satisfy the given equation:
Then, \(6\left(\frac{2}{3}\right)^2 - \frac{2}{3} - k = 0\)
\(\Rightarrow 6 \times \frac{4}{9} - \frac{2}{3} - k = 0\)
\(\Rightarrow \frac{8}{3} - \frac{2}{3} - k = 0\)
\(\Rightarrow k = 2\)
Hence, the value of \(k\) is 2.

Question. For what values of '\(a\)', does the quadratic equation \(x^2 - ax + 1 = 0\) not have real roots? 
Answer: Given quadratic equation is \(x^2 - ax + 1 = 0\)
On comparing the given equation with \(Ax^2 + Bx + C = 0\), we get:
\(A = 1, B = -a, C = 1\)
For real roots, \(D > 0\)
\(B^2 - 4AC > 0\)
i.e. \((-a)^2 - 4 \times 1 \times 1 = 0\)
\(a^2 > 4\)
or \(a > \sqrt{4}\)
or \(-2 > a > 2\)
Hence, the value of '\(a\)' lies between -2 and 2.

Question. If \(x = 3\) is one root of the quadratic equation \(x^2 - 2kx - 6 = 0\), then find the value of \(k\).
Answer: \(x^2 - 2kx - 6 = 0\). Let \(\alpha\) be other root.
Product \( = \frac{c}{a} = \frac{-6}{1} = -6\).
\(3 \times \alpha = -6 \Rightarrow \alpha = -2\).
Sum \( = \frac{-b}{a} = \frac{-(-2k)}{1} = 2k\).
\(\Rightarrow 3 + (-2) = 2k\)
\(1 = 2k\), \(k = \frac{1}{2}\)
Value of \(k\) is \(1/2\). [CBSE Topper 2018]

Question. Find the value of \(k\) for which the roots of the quadratic equation \(2x^2 + kx + 8 = 0\) will have equal value. 
Answer: Given: quadratic equation is \(2x^2 + kx + 8 = 0\)
For roots of the equation to be equal,
\(D = 0\)
i.e., \(b^2 - 4ac = 0\)
Here, \(a = 2, b = k\) and \(c = 8\)
\(k^2 - 4 \times 2 \times 8 = 0\)
\(k^2 = 64\)
\(k = \pm 8\)
Hence, the value of \(k\) is 8 or -8.

SHORT ANSWER (SA-I) Type Questions

Question. For what positive values of \(k\), does the quadratic equation \(3x^2 - kx + 3 = 0\) not have real roots? 
Answer: Given: quadratic equation \(3x^2 - kx + 3 = 0\), has no real roots.
On comparing the given equation with \(ax^2 + bx + c = 0\), we have:
\(a = 3, b = -k, c = 3\)
Then, discriminant,
\(D = b^2 - 4ac\)
\(= (-k)^2 - 4 \times 3 \times 3\)
\(= k^2 - 36\)
But for no real roots, \(D < 0\)
Then \(k^2 - 36 < 0\)
\(k^2 < 36\)
\(k < \pm 6\)
\(k > -6\) or \(k < 6\)
Hence, the value of \(k < 6\) (positive value) for no real roots.

Question. Solve for \(x\) : \(6x^2 + 11x + 3 = 0\) 
Answer: \(6x^2 + 11x + 3 = 0\)
\(6x^2 + 9x + 2x + 3 = 0\)
\(3x (2x + 3) + 1 (2x + 3) = 0\)
\((2x + 3) (3x + 1) = 0\)
\(2x + 3 = 0\) or \(3x + 1 = 0\)
i.e., \(x = -\frac{3}{2}\) or \(x = -\frac{1}{3}\).

Question. Solve for \(x\) : \(8x^2 - 2x - 3 = 0\) 
Answer: \(8x^2 - 2x - 3 = 0\)
\(8x^2 - 6x + 4x - 3 = 0\)
\(2x(4x - 3) + 1(4x - 3) = 0\)
\((4x - 3) (2x + 1) = 0\)
\(4x - 3 = 0\) or \(2x + 1 = 0\)
i.e., \(x = \frac{3}{4}\) or \(-\frac{1}{2}\).

Question. Solve the following quadratic equation : \(6a^2x^2 - 7abx - 3b^2 = 0\) 
Answer: Given: quadratic equation is :
\(6a^2x^2 - 7abx - 3b^2 = 0\)
On comparing the given equation with \(AX^2 + BX + C = 0\), we get:
\(A = 6a^2, B = -7ab, C = -3b^2\)
Then, discriminants,
\(D = \sqrt{B^2 - 4AC}\)
\(= \sqrt{(-7ab)^2 - 4 \times 6a^2 \times (-3b^2)}\)
\(= \sqrt{49a^2b^2 + 72a^2b^2}\)
\(= \sqrt{121a^2b^2}\)
\(= 11ab\)
Then, roots of the equation, \(x = \frac{-B \pm D}{2A}\)
\(= \frac{-(-7ab) \pm 11ab}{2 \times 6a^2}\)
\(= \frac{18ab}{12a^2}\) and \(\frac{-4ab}{12a^2}\)
\(= \frac{3b}{2a}\) and \(\frac{-b}{3a}\)
Hence, the roots of the given equation are: \(\frac{3b}{2a}\) and \(\frac{-b}{3a}\).

Question. Solve for \(x\) : \(\sqrt{3}x^2 + 10x - 8\sqrt{3} = 0\) 
Answer: Given: \(\sqrt{3}x^2 + 10x - 8\sqrt{3} = 0\)
On comparing the above equation with \(ax^2 + bx + c = 0\),
we get : \(a = \sqrt{3}, b = 10\) and \(c = -8\sqrt{3}\)
Then, discriminant, \(D = b^2 - 4ac\)
\(= (10)^2 - 4 \times \sqrt{3} \times (-8\sqrt{3})\)
\(= 100 + 96\)
\(= 196\)
Roots of equation, \(x = \frac{-b \pm \sqrt{D}}{2a}\)
\(= \frac{-10 \pm \sqrt{196}}{2 \times \sqrt{3}}\)
\(= \frac{-10 \pm 14}{2\sqrt{3}}\)
\(= \frac{4}{2\sqrt{3}}\) or \(-\frac{24}{2\sqrt{3}}\)
\(= \frac{2}{\sqrt{3}}\) or \(-\frac{12}{\sqrt{3}}\)
\(= \frac{2\sqrt{3}}{3}\) or \(-4\sqrt{3}\)
Hence, the roots of the given equation are \(\frac{2\sqrt{3}}{3}\) and \(-4\sqrt{3}\).

Question. A quadratic equation with integral coefficient has integral roots. Justify your answer. 
Answer: No, the given statement is not always true.
Consider the quadratic equation \(8x^2 - 2x - 1 = 0\)
By splitting the middle term,
\(8x^2 - 4x + 2x - 1 = 0\)
\(4x(2x - 1) + 1(2x - 1) = 0\)
\((4x + 1)(2x - 1) = 0\)
If \(4x + 1 = 0 \Rightarrow x = -1/4\)
\(2x - 1 = 0 \Rightarrow x = 1/2\)
So, the given equation has integral coefficients but no integral roots.
Hence, the given statement is false.

Question. Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.
Answer: Yes, there exists a quadratic equation whose coefficients are rational but both of its roots are irrational.
Consider the quadratic equation \(x^2 - 6x + 7 = 0\)
Here, \(D = b^2 - 4ac = (-6)^2 - 4(1)(7)\)
\(\Rightarrow D = 36 - 28 = 8\)
Since, discriminant is not a perfect square, therefore it will have irrational roots.
The roots will be \(\frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{8}}{2} = \frac{6 \pm 2\sqrt{2}}{2}\).
The roots will be \(3 \pm \sqrt{2}\)
i.e. \(3 + \sqrt{2}\) and \(3 - \sqrt{2}\), are both irrational.

Question. Solve for \(x\) : \(\frac{x+3}{x+2} = \frac{3x-7}{2x-3}, x \neq -2, \frac{3}{2}\) 
Answer: \(\Rightarrow (x + 3) (2x - 3) = (x + 2) (3x - 7)\)
\(\Rightarrow 2x^2 + 6x - 3x - 9 = 3x^2 + 6x - 7x - 14\)
\(\Rightarrow 2x^2 - 3x^2 + 3x + x - 9 + 14 = 0\)
\(\Rightarrow -x^2 + 4x + 5 = 0\)
\(\Rightarrow x^2 - 4x - 5 = 0\)
\(\Rightarrow x^2 - 5x + x - 5 = 0\)
(on splitting the middle term)
\(\Rightarrow x(x - 5) + 1(x - 5) = 0\)
\(\Rightarrow (x + 1) (x - 5) = 0\)
\(\Rightarrow x = -1, 5\)
Hence, the values of \(x\) are -1 and 5.

Question. Find the roots of the quadratic equation \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\). 
Answer: Given, quadratic equation is \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\).
On comparing the above equation with \(ax^2 + bx + c = 0\)
we get \(a = \sqrt{2}, b = 7, c = 5\sqrt{2}\)
Then, discriminant, \(D = b^2 - 4ac\)
\(= (7)^2 - 4 \times \sqrt{2} \times 5\sqrt{2}\)
\(= 49 - 40 = 9\)
Now, \(x = \frac{-b \pm \sqrt{D}}{2a}\)
\(= \frac{-7 \pm \sqrt{9}}{2\sqrt{2}} = \frac{-7 \pm 3}{2\sqrt{2}}\)
\(\therefore x = \frac{-4}{2\sqrt{2}}\) and \(\frac{-10}{2\sqrt{2}}\)
\(= \frac{-2}{\sqrt{2}}\) and \(\frac{-5}{\sqrt{2}}\)
Hence, the roots of the given equation is \(-\sqrt{2}\) and \(-\frac{5}{\sqrt{2}}\).

Question. If \(b = 0\) and \(c < 0\), is it true that the roots of \(x^2 + bx + c = 0\) are numerically equal and opposite in sign? Justify.
Answer: It is given that \(b = 0\) and \(c < 0\).
The given quadratic equation is: \(x^2 + bx + c = 0\)
On putting \(b = 0\) in this equation, we get
\(x^2 + 0.x + c = 0\)
\(x^2 + c = 0\)
\(\Rightarrow x^2 = -c\)
Here, \(c < 0 \Rightarrow -c > 0\)
\(\Rightarrow x = \pm \sqrt{-c}\)
Hence, the roots of \(x^2 + bx + c = 0\) are numerically equal and opposite in sign.

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, quadratic equation is: \(x^2 + 2xk + (k^2 - k + 2) = 0\)
On comparing the quadratic equation, with \(ax^2 + bx + c = 0\), we get:
\(a = 1, b = 2k, c = k^2 - k + 2\)
Since, the roots of the above equation are real and equal.
\(\therefore\) Discriminant, \(D = 0\)
i.e., \(b^2 - 4ac = 0\)
\((2k)^2 - 4 \times 1 \times (k^2 - k + 2) = 0\)
\(4k^2 - 4k^2 + 4k - 8 = 0\)
\(4k - 8 = 0\)
\(k = 2\)
Hence, the value of \(k\) is 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: Since \(x = \frac{2}{3}\) and \(x = -3\) are the roots of the quadratic equation \(ax^2 + 7x + b = 0\)
Now, sum of roots: \(\frac{2}{3} + (-3) = -\frac{7}{a}\)
\(\Rightarrow -\frac{7}{3} = -\frac{7}{a} \Rightarrow a = 3\).
Product of roots: \(\frac{2}{3} \times (-3) = \frac{b}{a}\)
\(\Rightarrow -2 = \frac{b}{3}\) [\(\because a = 3\)]
\(\Rightarrow b = -6\)
Hence, the values of \(a\) and \(b\) are 3 and -6 respectively.

Question. If \(a\) and \(b\) are the roots of the equation \(x^2 + ax - b = 0\), then find \(a\) and \(b\).
Answer: \(x^2 + ax - b = 0\)
Sum of the roots \(= a + b = \frac{-\text{Coefficient of } x}{\text{Coefficient of } x^2} = -a\)
Product of roots \(= ab = \frac{\text{constant term}}{\text{Coefficient of } x^2} = -b\)
So, \(a + b = -a\)
\(b = -2a\)
and, \(ab = -b\)
\(a = -1\)
Putting the value of \(a\), we get
\(b = -2 \times (-1) = 2\)
Hence, \(a = -1\) and \(b = 2\).