250+ TOP MCQs on Quadratic Equations & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Quadratic Equations – 1”.

1. Roots of a quadratic equation are real when discriminant is ______________
a) zero
b) greater than zero
c) less than zero
d) greater than or equal to zero
Answer: d
Clarification: For a quadratic equation, ax²+bx+c = 0, discriminant is b²-4ac.
Roots are (frac{-b±sqrt{b^2-4ac}}{2a}). For real roots, radical must be non-negative i.e. discriminant should be greater than or equal to zero.

2. Roots of a quadratic equation are imaginary when discriminant is ______________
a) zero
b) greater than zero
c) less than zero
d) greater than or equal to zero
Answer: c
Clarification: For a quadratic equation, ax²+bx+c = 0, discriminant is b²-4ac.
Roots are (frac{-b±sqrt{b^2-4ac}}{2a}). For imaginary roots, radical is negative i.e. discriminant should be less than zero.

3. Solve x²+1 = 0.
a) x=1, -1
b) x=i, -i
c) x=-1
d) x=i
Answer: b
Clarification: x²+1 = 0
=>x² = -1 => x = ±(sqrt{-1}) = ±i.

4. Solve 2x² + x + 1 = 0.
a) (frac{-1±isqrt{7}}{4})
b) (frac{1±isqrt{7}}{4})
c) (frac{1±sqrt{7}}{4})
d) (frac{-1±sqrt{7}}{4})
Answer: a
Clarification: 2x² + x + 1 = 0
D=1²-4*2*1 = 1-8 = -7 ≤ 0.
Since D ≤ 0, imaginary roots are there.
=>x = (frac{-1±sqrt{1^2-4 ˙ 2.1}}{2.2} = frac{-1±isqrt{7}}{4}).

5. Solve – x² + x – 2 = 0.
a) (frac{-1±isqrt{7}}{2})
b) (frac{1±isqrt{7}}{2})
c) (frac{1±sqrt{7}}{2})
d) (frac{-1±sqrt{7}}{2})
Answer: b
Clarification: – x² + x – 2 = 0
=>x²-x+2 = 0
D=(-1)²-4*1*2 = 1-8 = -7 ≤ 0.
Since D ≤ 0, imaginary roots are there.
=>x = (frac{1±sqrt{D}}{2.1} = frac{1±isqrt{7}}{2}).

6. Solve 2x²+√2x+2= 0.
a) (frac{-1±isqrt{7}}{2sqrt{2}})
b) (frac{1±isqrt{7}}{2sqrt{2}})
c) (frac{1±sqrt{7}}{2sqrt{2}})
d) (frac{-1±sqrt{7}}{2sqrt{2}})
Answer: a
Clarification: 2x²+√2x+2 = 0
=>D=((sqrt{2}))² – 4.2.2 = 2-16 = -14.
Since D ≤ 0, imaginary roots are there.
=>x = (frac{-sqrt{2}±sqrt{D}}{2.2} = frac{-sqrt{2}±isqrt{14}}{4} = frac{-1±isqrt{7}}{2sqrt{2}}).

7. Solve (sqrt{3})x² + x + (sqrt{3}) = 0
a) (frac{-1±isqrt{11}}{6sqrt{3}})
b) (frac{1±isqrt{11}}{6sqrt{3}})
c) (frac{1±sqrt{11}}{6sqrt{3}})
d) (frac{-1±sqrt{11}}{6sqrt{3}})
Answer: a
Clarification: (sqrt{3})x² + x + (sqrt{3}) = 0
=>3x² + √3x + 3 = 0
=>D = (√3)² – 4.3.3 = 3-36 = -33.
Since D ≤ 0, imaginary roots are there.
=>x = (frac{-sqrt{3}±isqrt{33}}{2.3} = frac{-1±isqrt{11}}{6sqrt{3}}).

8. Solve (sqrt{3}x^2 – sqrt{2} x + 3sqrt{3}) = 0
a) (frac{-1±isqrt{17}}{2})
b) (frac{1±isqrt{17}}{2})
c) (frac{1±sqrt{17}}{2})
d) (frac{-1±sqrt{17}}{2})
Answer: b
Clarification: (sqrt{3}x^2 – sqrt{2} x + 3sqrt{3}) = 0
=>3x² – (sqrt{6})x + 9 = 0
=>D=(-(sqrt{6}))² – 4.3.9 = 6-108 = -102.
Since D ≤ 0, imaginary roots are there.
=>x = (frac{sqrt{6}±isqrt{102}}{2.3} = frac{1±isqrt{17}}{2}).