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.
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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 = \(\frac{1±\sqrt{D}}{2.1} = \frac{1±i\sqrt{7}}{2}\).
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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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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter