Mathematics Questions and Answers – Quadratic Equations – 1

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses 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
View Answer

Answer: d
Explanation: 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
View Answer

Answer: c
Explanation: 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
View Answer

Answer: b
Explanation: x2+1 = 0
=>x2 = -1 => x = ±\(\sqrt{-1}\) = ±i.
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4. Solve 2x2 + x + 1 = 0.
a) \(\frac{-1±i\sqrt{7}}{4}\)
b) \(\frac{1±i\sqrt{7}}{4}\)
c) \(\frac{1±\sqrt{7}}{4}\)
d) \(\frac{-1±\sqrt{7}}{4}\)
View Answer

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

5. Solve – x2 + x – 2 = 0.
a) \(\frac{-1±i\sqrt{7}}{2}\)
b) \(\frac{1±i\sqrt{7}}{2}\)
c) \(\frac{1±\sqrt{7}}{2}\)
d) \(\frac{-1±\sqrt{7}}{2}\)
View Answer

Answer: b
Explanation: – 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 = \(\frac{1±\sqrt{D}}{2.1} = \frac{1±i\sqrt{7}}{2}\).

6. Solve 2x2+√2x+2= 0.
a) \(\frac{-1±i\sqrt{7}}{2\sqrt{2}}\)
b) \(\frac{1±i\sqrt{7}}{2\sqrt{2}}\)
c) \(\frac{1±\sqrt{7}}{2\sqrt{2}}\)
d) \(\frac{-1±\sqrt{7}}{2\sqrt{2}}\)
View Answer

Answer: a
Explanation: 2x2+√2x+2 = 0
=>D=(\(\sqrt{2}\))2 – 4.2.2 = 2-16 = -14.
Since D ≤ 0, imaginary roots are there.
=>x = \(\frac{-\sqrt{2}±\sqrt{D}}{2.2} = \frac{-\sqrt{2}±i\sqrt{14}}{4} = \frac{-1±i\sqrt{7}}{2\sqrt{2}}\).

7. Solve \(\sqrt{3}\)x2 + x + \(\sqrt{3}\) = 0
a) \(\frac{-1±i\sqrt{11}}{6\sqrt{3}}\)
b) \(\frac{1±i\sqrt{11}}{6\sqrt{3}}\)
c) \(\frac{1±\sqrt{11}}{6\sqrt{3}}\)
d) \(\frac{-1±\sqrt{11}}{6\sqrt{3}}\)
View Answer

Answer: a
Explanation: \(\sqrt{3}\)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 = \(\frac{-\sqrt{3}±i\sqrt{33}}{2.3} = \frac{-1±i\sqrt{11}}{6\sqrt{3}}\).
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8. Solve \(\sqrt{3}x^2 – \sqrt{2} x + 3\sqrt{3}\) = 0
a) \(\frac{-1±i\sqrt{17}}{2}\)
b) \(\frac{1±i\sqrt{17}}{2}\)
c) \(\frac{1±\sqrt{17}}{2}\)
d) \(\frac{-1±\sqrt{17}}{2}\)
View Answer

Answer: b
Explanation: \(\sqrt{3}x^2 – \sqrt{2} x + 3\sqrt{3}\) = 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 = \(\frac{\sqrt{6}±i\sqrt{102}}{2.3} = \frac{1±i\sqrt{17}}{2}\).

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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