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

 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) x2+3x=(1)(13x)
x2+3x=1+3x

x2+1=0

It is also quadratic equation.


(c) (x+2)(x1)=x22x3

x2x+2x2=x22x3

x2x2+x+2x2+3=03x+1=0

It is not a quadratic equation.


(d) x3x2+2x+1=(x+1)3

=x3+3x2+3x+1

x3x2+2x+1

3x2+x22x1+3x+1=0

4x2+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=2x(x21)

(c) x2+3x+1=(x2)2

(d) 8(x2)3=(2x1)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=2x(x21)

x3+6x2+12x+8=2x32x

x3+6x2+12x+82x3+2x=0

x3+6x2+14x+8=0

It is not a quadratic equation.


(c) x2+3x+1=(x2)2

x2+3x+1=x24x+4

⇒3x+1+4 x-4=0 

⇒7x-3=0

It is not a quadratic equation.


(d) 8(x2)3=(2x1)3+3

8(x36x2+12x8)

=8x312x2+6x1+3

8x348x2+96x648x3+12x26x+13=0

36x2+90x66=0

It is a quadratic equation

Ans : (d)


Question 3

Which of the following equations has 2 as a root ?

(a) x24x+5=0

(b) x2+3x12=0

(c) 2x27x+6=0

(d) 3x26x2=0

Sol :

(a) x24x+5=0

(2)24x2+5=0

⇒ 4 – 8 + 5 = 0

⇒ 9 – 8 ≠ 0

2 is not its root.


(b) x2+3x12=0
(2)23×212=0
⇒4-6-12=4-18=-14
∴ 2 is not its roots.

(c) 2x27x+6=0
2(2)27×2+6=0
⇒ 8-14+6=0 $
⇒0=0
∴ 2 is its root

(d) 3x26x2=0
3(2)26×22=0
⇒12-12-2=0 
⇒12-14=0
∴ 2 is not its root.

Ans : (c)


Question 4

If 12 is a root of the equation x2+kx54=0 then the value of k is

(a) 2

(b) – 2

(c) 14

(d) 12

Sol :

12 is a root of the equation

x2+kx54=0

Substituting the value of x=12 in the equation 

(12)2+k×1254=0

14+k254=0

k21=0

⇒k=1×2=2

∴k=2

Ans (a)


Question 5

If 12 is a root of the quadratic equation 4x24kx+k+5=0 then the value of k is

(a) – 6
(b) – 3
(c) 3
(d) 6

Sol :

12 is a root of the equation 

4x24kx+k+5=0

Substituting the value of x=12 in the equation

4(12)24×k×12+k+5=0

1-2k+k+5=0

-k+6=0

k=6

Ans (d)


Question 6

The roots of the equation x23x10=0 are

(a) 2,- 5

(b) – 2, 5

(c) 2, 5

(d) – 2, – 5

Sol :
x=(3)±(3)24×1×(10)2×1

=3±9+402=3±492=3+72

x=3+72=5 and x=372=42=2

x = 5, – 2 or – 2, 5 

Ans (b)


Question 7

If one root of a quadratic equation with rational coefficients is 352, then the other 

(a) 352

(b) 3+52

(c) 3+52

(d) 3+52

Sol :

One root of a quadratic equation is 352 then other root will be 3+52

Ans (c)


Question 8

If the equation 2x25x+(k+3)=0 has equal roots then the value of k is

(a) g8

(b) g8

(c) 18

(d) 18

Sol :

2x25x+(k+3)=0

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

=25-8(k+3)

∴ Roots are equal. 

b24ac=0

∴ 25-8(k+3)=0

⇒25-8k-24=0

⇒1-8k=0 

⇒8 k=1

k=18

Ans (c)


Question 9

The value(s) of k for which the quadratic equation 2x2kx+k=0 has equal roots is (are)

(a) 0 only

(b) 4

(c) 8 only

(d) 0, 8

Sol :

2x2kx+k=0

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

b24ac=(k)24×2×k

=k28k

∴ Roots are equal. b24ac=0

k28k=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 3x2kx+2k=0 roots, then the the value(s) of k is (are)

(a) 6

(b) 0 Only

(c) 24 only

(d) 0

Sol :

3x2kx+2k=0

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

b24ac=(k)24×3×2k

=k224k

∴ Roots are equal. b24ac=0

k224k=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}x22(k1)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

b24ac=[2(k1)]24(k+1)(1)
=4(k22k+1)4k4
=4k28k+44k4
=4k212k
∵ Roots are equal. b24ac=0
4k212k=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<92
(b)p 92
(c) p>92
(d) p92
Sol :
2x² – 6x + p = 0
Here, a = 2, b = -6, c = p
b24ac=(6)24×2×p
=36-8 p
∵ Roots are real and unequal. 
b24ac>0

⇒ 36-8p>0

⇒ 36-8p>0

⇒ 36>8p

⇒ 368>p

⇒ p<368

p<92

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

b24ac=(5)24×2×1
=5-8=-3

b24ac<0

∵ It has no real roots.


Question 14

Which of the following equations has two distinct real roots ?

(a) 2x232x+94=0
(b) x2+x5=0
(c) x2+3x+22=0
(d) 5x23x+1=0

Sol :

(a) 2x232x+94=0

b24ac=(32)24×2×94=1818=0

∵ Roots are real and equal.


(b) x2+x5=0

b24ac=(1)24×1×(5)

=1+20=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)

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