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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

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

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

Explanation: x2+1 = 0
=>x2 = -1 => x = ±$$\sqrt{-1}$$ = ±i.

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) $$\frac{-1±\sqrt{7}}{4}$$

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) $$\frac{-1±\sqrt{7}}{2}$$

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) $$\frac{-1±\sqrt{7}}{2\sqrt{2}}$$

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) $$\frac{-1±\sqrt{11}}{6\sqrt{3}}$$

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 = $$\frac{-\sqrt{3}±i\sqrt{33}}{2.3} = \frac{-1±i\sqrt{11}}{6\sqrt{3}}$$.

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) $$\frac{-1±\sqrt{17}}{2}$$

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.