CBSE Class 10 Mathematics Quadratic Equations MCQs Set M

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) ± 4
Explanation:
The equation \( 2x^2 + kx + 2 = 0 \) has equal roots, when \( D = (k)^2 – 4(2)(2) = 0 \)
(\( \because D = b^2 – 4ac \))
\( k^2 – 16 = 0 \)
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) k < 4
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) k < 16
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) \( x^3 – x^2 = (x – 1)^3 \)
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) \( (\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 - 5x \)
Explanation:
It is given that
(a) \( 2(x – 1)^2 = 4x^2 – 2x + 1 \)
\( \Rightarrow 2(x^2 + 1 – 2x) = 4x^2 – 2x + 1 \)
\( \Rightarrow 2x^2 + x^2 – 4x = 4x^2 – 2x + 1 \)
\( \Rightarrow 2x^2 + 2x – 1 = 0 \)
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) \( (x^2 + 2x)^2 = x^4 + 3 + 4x^3 \)
\( x^4 + 4x^2 + 4x^3 = x^4 + 3 + 4x^3 \)
\( 4x^2 – 3 = 0 \)
This represents 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) k < 8
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) \( 2x^2 – 7x + 6 = 0 \)
Explanation:
(a) Putting the value of \( x = 2 \) in \( x^2 – 4x + 5 = 0 \), we get
\( (2)^2 – 4(2) + 5 = 0 \)
\( \Rightarrow 4 – 8 + 5 = 0 \)
\( \Rightarrow 1 \neq 0 \)
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
\( (2)^2 + 3(2) – 12 = 0 \)
\( \Rightarrow 4 + 6 – 12 = 0 \)
\( \Rightarrow –2 \neq 0 \)
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
\( 2(2)^2 – 7(2) + 6 = 0 \)
\( \Rightarrow 8 – 14 + 6 = 0 \)
\( \Rightarrow 0 = 0 \)
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
\( 3(2)^2 – 6(2) – 2 = 0 \)
\( \Rightarrow 12 – 12 – 2 = 0 \)
\( \Rightarrow –2 \neq 0 \)
So, \( x = 2 \) is not the root of the equation \( 3x^2 – 6x – 2 = 0 \).

Question. The positive root of \( \sqrt{3x^2 + 6} = 9 \) is
(a) 2
(b) 1
(c) 4
(d) 3
Answer: (b) 1
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) \( –x^2 + 3x – 3 = 0 \)
Explanation:
We know that sum of the roots = \( -\frac{b}{a} \)
On comparing the given equations with \( ax^2 + bx + c = 0 \):
(a) \( 2x^2 – 3x + 6 = 0 \)
\( a = 2, b = –3, c = 6 \)
\( \Rightarrow \) Sum of its roots = \( -\frac{b}{a} = -\frac{(-3)}{2} = \frac{3}{2} \)
(b) \( –x^2 + 3x – 3 = 0 \)
\( a = –1, b = 3, c = –3 \)
\( \Rightarrow \) Sum of its roots = \( -\frac{b}{a} = -\left( \frac{3}{-1} \right) = 3 \)
(c) \( \sqrt{2}x^2 - \frac{3}{\sqrt{2}}x + 1 = 0 \)
\( \Rightarrow a = \sqrt{2}, b = -\frac{3}{\sqrt{2}}, c = 1 \)
\( \Rightarrow \) Sum of its roots = \( -\frac{b}{a} = -\frac{-3/\sqrt{2}}{\sqrt{2}} = \frac{3}{2} \)
(d) \( 3x^2 – 3x + 3 = 0 \)
\( \Rightarrow a = 3, b = –3 \) and \( c = 3 \)
\( \Rightarrow \) 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) yes
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) k = 0, 6
Explanation:
We have \( kx(x – 2) + 6 = 0 \)
\( kx^2 – 2kx + 6 = 0 \)
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) \( \frac{1}{2}, -1 \)
Answer: (a) \( -\frac{1}{2}, 1 \)
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) no real roots
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, \( D = b^2 – 4ac \)
\( = (– \sqrt{5})^2 – 4(2)(1) \)
\( D = 5 – 8 = –3 \)
\( \Rightarrow D = –3 < 0 \)
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) \( x^2 + x – 5 = 0 \)
Explanation:
We know that if \( D = b^2 – 4ac > 0 \) for the quadratic equation \( ax^2 + bx + c = 0 \), then its roots are real and distinct.
(a) The given equation is:
\( 2x^2 – 3\sqrt{2}x + \frac{9}{4} = 0 \)
\( \Rightarrow a = 2, b = –3\sqrt{2}, c = \frac{9}{4} \)
\( \Rightarrow D = b^2 – 4ac = (–3\sqrt{2})^2 – 4(2)\left(\frac{9}{4}\right) \)
\( = 18 – 18 = 0 \)
So, the equation has real and equal roots.
(b) The given equation is:
\( x^2 + x – 5 = 0 \)
\( \Rightarrow a = 1, b = 1, c = –5 \)
\( \Rightarrow D = b^2 – 4ac = (1)^2 – 4(1)(–5) = 1 + 20 = 21 > 0 \)
So, the equation has two distinct real roots.
(c) The given equation is:
\( x^2+ 3x + 2\sqrt{2} = 0 \)
\( \Rightarrow a = 1, b = 3, c = 2\sqrt{2} \)
\( \Rightarrow D = b^2 – 4ac = (3)^2 – 4(1)(2\sqrt{2}) = 9 – 8\sqrt{2} < 0 \)
So, the equation has no real roots.
(d) The given equation is:
\( 5x^2 – 3x + 1 = 0 \)
\( \Rightarrow a = 5, b = –3, c = 1 \)
\( \Rightarrow D = b^2 – 4ac = (–3)^2 – 4(5)(1) = 9 – 20 = –11 < 0 \)
So, the equation has 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) no real root
Explanation:
The given equation is:
\( (x^2 + 1)^2 – x^2 = 0 \)
\( \Rightarrow x^4 + 1 + 2x^2 – x^2 = 0 \)
[\( \because (a + b)^2 = a^2 + b^2 + 2ab \)]
\( \Rightarrow x^4 + x^2 + 1 = 0 \)
Let \( x^2 = y \)
\( (x^2)^2 + x^2 + 1 = 0 \)
\( \Rightarrow y^2 + y + 1 = 0 \)
On comparing with \( ay^2 + by + c = 0 \), we get \( a = 1, b = 1, c = 1 \)
\( \Rightarrow D = b^2 – 4ac = (1)^2 – 4(1)(1) = –3 < 0 \)
As \( D < 0 \), thus, we can say equation has no real roots for \( y^2 + y + 1 = 0 \)
i.e. \( x^4 + x^2 + 1 = 0 \) or \( (x^2 + 1)^2 – x^2 = 0 \) has no real roots.

Fill in the Blanks

Fill in the blanks/tables with suitable information:

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 \( p^2 – 4(2)(3) = 0 \)
i.e. when \( p^2 – 24 = 0 \), or \( p = \sqrt{24} \) or \( p = \pm 2\sqrt{6} \)

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 = \( -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{-4}{1} = 4 \)
Product of roots = \( \frac{\text{constant term}}{\text{Coefficient of } x^2} = \frac{2}{1} = 2 \)
Thus sum is twice the product.

Question. The quadratic equation \( 2x^2 + x + 4 \) has .................... real roots.
Answer: No
Explanation:
Here, \( b^2 – 4ac = 1 – 32 \), 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:
The sum roots of the equation \( 2x^2 + 14x + 24 = 0 \) is \( \frac{-14}{2} \), i.e., –7