ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable MCQs

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15

 MCQs

Question 1

Which of the following is not a quadratic equation ?

(a) (x+2)²=2(x+3)

(b) x²+3x=(–1) (1–3x)

(c) (x+2)(x–1)=x²–2x–3

(d) x³–x²+2x+1=(x+1)3

Sol :

(a) (x + 2)² = 2(x + 3)

⇒ x² + 4x + 4 = 2x + 6

⇒ x² + 4x – 2x + 4 – 6 = 0

⇒ x² + 2x – 2

It is a quadratic equation.

(b) $x^{2}+3 x=(-1)(1-3 x) $

$\Rightarrow x^{2}+3 x=-1+3 x$

$\Rightarrow x^{2}+1=0$

It is also quadratic equation.

(c) $(x+2)(x-1)=x^{2}-2 x-3$

$x^{2}-x+2 x-2=x^{2}-2 x-3$

$x^{2}-x^{2}+x+2 x-2+3=0 \Rightarrow 3 x+1=0$

It is not a quadratic equation.

(d) $x^{3}-x^{2}+2 x+1=(x+1)^{3}$

$=x^{3}+3 x^{2}+3 x+1$

$x^{3}-x^{2}+2 x+1$

$3 x^{2}+x^{2}-2 x-1+3 x+1=0$

$\Rightarrow 4 x^{2}+x=0$

It is a quadratic equation 

Ans : (c)

Question 2

Which of the following is a quadratic equation ?

(a) (x – 2) (x + 1) = (x – 1) (x – 3)

(b) $(x+2)^{3}=2 x\left(x^{2}-1\right)$

(c) $x^{2}+3 x+1=(x-2)^{2}$

(d) $8(x-2)^{3}=(2 x-1)^{3}+3$

Sol :

(a) (x – 2) (x + 1) = (x – 1) (x – 3)
⇒ x2 + x – 2x – 2 = x2 – 3x – x + 3
⇒ 3x + x – 2x + x = 3 + 2
⇒ 3x = 5
It is not a quadratic equation.

(b) $(x+2)^{3}=2 x\left(x^{2}-1\right)$

$x^{3}+6 x^{2}+12 x+8=2 x^{3}-2 x$

$x^{3}+6 x^{2}+12 x+8-2 x^{3}+2 x=0$

$-x^{3}+6 x^{2}+14 x+8=0$

It is not a quadratic equation.

(c) $x^{2}+3 x+1=(x-2)^{2}$

$x^{2}+3 x+1=x^{2}-4 x+4$

⇒3x+1+4 x-4=0 

⇒7x-3=0

It is not a quadratic equation.

(d) $8(x-2)^{3}=(2 x-1)^{3}+3$

$8\left(x^{3}-6 x^{2}+12 x-8\right)$

$=8 x^{3}-12 x^{2}+6 x-1+3$

$8 x^{3}-48 x^{2}+96 x-64-8 x^{3}+12 x^{2}-6 x+1-3=0$

$-36 x^{2}+90 x-66=0$

It is a quadratic equation

Ans : (d)

Question 3

Which of the following equations has 2 as a root ?

(a) $x^{2}-4 x+5=0$

(b) $x^{2}+3 x-12=0$

(c) $2 x^{2}-7 x+6=0$

(d) $3 x^{2}-6 x-2=0$

Sol :

(a) $x^{2}-4 x+5=0$

$\Rightarrow(2)^{2}-4 x^{2}+5=0$

⇒ 4 – 8 + 5 = 0

⇒ 9 – 8 ≠ 0

2 is not its root.

(b) $x^{2}+3 x-12=0$

$ \Rightarrow(2)^{2}-3 \times 2-12=0$

⇒4-6-12=4-18=-14

∴ 2 is not its roots.

(c) $2 x^{2}-7 x+6=0$

$ \Rightarrow 2(2)^{2}-7 \times 2+6=0$

⇒ 8-14+6=0 $

⇒0=0

∴ 2 is its root

(d) $3 x^{2}-6 x-2=0$

$ \Rightarrow 3(2)^{2}-6 \times 2-2=0$

⇒12-12-2=0 

⇒12-14=0

∴ 2 is not its root.

Ans : (c)

Question 4

If $\frac{1}{2}$ is a root of the equation $x^{2}+k x-\frac{5}{4}=0$ then the value of k is

(a) 2

(b) – 2

(c) $\frac{1}{4}$

(d) $\frac{1}{2}$

Sol :

$\frac{1}{2}$ is a root of the equation

$x^{2}+k x-\frac{5}{4}=0$

Substituting the value of $x=\frac{1}{2}$ in the equation 

$\left(\frac{1}{2}\right)^{2}+k \times \frac{1}{2}-\frac{5}{4}=0$

$\Rightarrow \frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0$

$\Rightarrow \frac{k}{2}-1=0$

⇒k=1×2=2

∴k=2

Ans (a)

Question 5

If $\frac{1}{2}$ is a root of the quadratic equation $4 x^{2}-4 k x+k+5=0$ then the value of k is

(a) – 6

(b) – 3

(c) 3

(d) 6

Sol :

$\frac{1}{2}$ is a root of the equation 

$4 x^{2}-4 k x+k+5=0$

Substituting the value of $x=\frac{1}{2}$ in the equation

$4\left(\frac{1}{2}\right)^{2}-4 \times k \times \frac{1}{2}+k+5=0$

1-2k+k+5=0

-k+6=0

k=6

Ans (d)

Question 6

The roots of the equation $x^{2}-3 x-10=0$ are

(a) 2,- 5

(b) – 2, 5

(c) 2, 5

(d) – 2, – 5

Sol :

$x=\frac{-(-3) \pm \sqrt{(-3)^{2}-4 \times 1 \times(-10)}}{2 \times 1}$

$=\frac{3 \pm \sqrt{9+40}}{2}=\frac{3 \pm \sqrt{49}}{2}=\frac{3+7}{2}$

∴$x=\frac{3+7}{2}=5$ and $x=\frac{3-7}{2}=\frac{-4}{2}=-2$

x = 5, – 2 or – 2, 5 

Ans (b)

Question 7

If one root of a quadratic equation with rational coefficients is $\frac{3-\sqrt{5}}{2}$, then the other 

(a) $\frac{-3-\sqrt{5}}{2}$

(b) $\frac{-3+\sqrt{5}}{2}$

(c) $\frac{3+\sqrt{5}}{2}$

(d) $\frac{\sqrt{3}+5}{2}$

Sol :

One root of a quadratic equation is $\frac{3-\sqrt{5}}{2}$ then other root will be $\frac{3+\sqrt{5}}{2}$

Ans (c)

Question 8

If the equation $2 x^{2}-5 x+(k+3)=0$ has equal roots then the value of k is

(a) $\frac{g}{8}$

(b) $-\frac{g}{8}$

(c) $\frac{1}{8}$

(d) $-\frac{1}{8}$

Sol :

$2 x^{2}-5 x+(k+3)=0$

a=2, b=-5, c=k+3

=25-8(k+3)

∴ Roots are equal. 

$\therefore b^{2}-4 a c=0$

∴ 25-8(k+3)=0

⇒25-8k-24=0

⇒1-8k=0 

⇒8 k=1

$\therefore k=\frac{1}{8}$

Ans (c)

Question 9

The value(s) of k for which the quadratic equation $2 x^{2}-k x+k=0$ has equal roots is (are)

(a) 0 only

(b) 4

(c) 8 only

(d) 0, 8

Sol :

$2 x^{2}-k x+k=0$

a=2, b=-k, c=k

$\therefore b^{2}-4 a c=(-k)^{2}-4 \times 2 \times k$

$\quad=k^{2}-8 k$

∴ Roots are equal. $\therefore b^{2}-4 a c=0$

$k^{2}-8 k=0$

⇒k(k-8)=0$

Either k=0

or k-8=0, then k=8

k=0,8

Ans (d)

Question 10

If the equation $3 x^{2}-k x+2 k=0$ roots, then the the value(s) of k is (are)

(a) 6

(b) 0 Only

(c) 24 only

(d) 0

Sol :

$3 x^{2}-k x+2 k=0$

Here, a=3, b=-k, c=2 k

$b^{2}-4 a c=(-k)^{2}-4 \times 3 \times 2 k$

$=k^{2}-24 k$

∴ Roots are equal. $\therefore b^{2}-4 a c=0$

$\therefore k^{2}-24 k=0$

⇒k(k-24)=0

Either k=0 

or k-24=0, then k=24

∴ k=0, 24 

Ans (d)

Question 11

If the equation $\{k+1\} x^{2}-2(k-1) x+1=0$ has equal roots, then the values of k are

(a) 1, 3

(b) 0, 3

(c) 0, 1

(d) 0, 1

Sol :

(k + 1)x² – 2(k – 1)x + 1 = 0

Here, a = k + 1, b = -2(k – 1), c = 1

$\therefore b^{2}-4 a c=[-2(k-1)]^{2}-4(k+1)(1)$

$\quad=4\left(k^{2}-2 k+1\right)-4 k-4$

$\quad=4 k^{2}-8 k+4-4 k-4$

$\quad=4 k^{2}-12 k$

∵ Roots are equal. $\therefore b^{2}-4 a c=0$

$\therefore 4 k^{2}-12 k=0$

⇒4 k(k-3)=0 

⇒ k(k-3)=0

Either k=0 or k-3=0, then k=3

k=0,3(b)

Question 12

If the equation 2x² – 6x + p = 0 has real and different roots, then the values ofp are given by

(a) $p<\frac{9}{2}$

(b)p $\leq \frac{9}{2}$

(c) $p>\frac{9}{2}$

(d) $p \geq \frac{9}{2}$

Sol :

2x² – 6x + p = 0

Here, a = 2, b = -6, c = p

$b^{2}-4 a c=(-6)^{2}-4 \times 2 \times p$

=36-8 p

∵ Roots are real and unequal. 

$\therefore b^{2}-4 a c>0$

⇒ 36-8p>0

⇒ 36-8p>0

⇒ 36>8p

⇒ $\frac{36}{8}>p$

⇒ $p<\frac{36}{8}$

⇒ $p<\frac{9}{2}$

Ans (a)

Question 13

The quadratic equation 2x² – √5x + 1 = 0 has

(a) two distinct real roots

(b) two equal real roots

(c) no real roots

(d) more than two real roots

Sol :

2x² – √5x + 1 = 0

Here, a = 2, b = -√5, c = 1

$b^{2}-4 a c=(-\sqrt{5})^{2}-4 \times 2 \times 1$

=5-8=-3

$\because b^{2}-4 a c<0$

∵ It has no real roots.

Question 14

Which of the following equations has two distinct real roots ?

(a) $2 x^{2}-3 \sqrt{2 x}+\frac{9}{4}=0$

(b) $x^{2}+x-5=0$

(c) $x^{2}+3 x+2 \sqrt{2}=0$

(d) $5 x^{2}-3 x+1=0$

Sol :

(a) $2 x^{2}-3 \sqrt{2} x+\frac{9}{4}=0$

$b^{2}-4 a c=(-3 \sqrt{2})^{2}-4 \times 2 \times \frac{9}{4}=18-18=0$

∵ Roots are real and equal.

(b) $x^{2}+x-5=0$

$b^{2}-4 a c=(1)^{2}-4 \times 1 \times(-5)$

$=1+20=\sqrt{21}>0$

Roots are real and distinct.

Ans (b)

Question 15

Which of the following equations has no real roots ?

(a) x² – 4x + 3√2 = 0

(b) x² + 4x – 3√2 = 0

(c) x² – 4x – 3√2 = 0

(d) 3x² + 4√3x + 4 = 0

Sol :

(a) x² – 4x + 3√2 = 0

b² – 4ac = ( -4)² – 4 × 1 × 3√2

= 16 – 12√2

= 16 – 12(1.4)

= 16 – 16.8

= -0.8

b² – 4ac < 0

Roots are not real.

Ans (a)