# GSEB Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3

Solve each of the following equations:
1. x2 + 3 = 0
2. 2x2 + x + 1 = 0
3. x2 + 3x + 9 = 0
4. – x2 + x – 2 = 0
5. x2 + 3x + 5 = 0
6. x2 – x + 2 = 0
7. $$\sqrt{2}$$x2 + x + $$\sqrt{2}$$ = 0
8. $$\sqrt{3}$$x2 – $$\sqrt{2}$$x + 3$$\sqrt{3}$$ = 0
9. x2 + x + $$\frac{1}{\sqrt{2}}$$ = 0
10. x2 + $$\frac{x}{\sqrt{2}}$$ + 1 = 0.
Solutions to questions 1 to 10:
1. x2 + 3 = 0 ⇒ x2 = – 3
∴ x = ±$$\sqrt{- 3}$$ = ±$$\sqrt{3i}$$.

2. 2x2 + x + 1 = 0.
Comparing with ax2 + bx + c = 0,
a = 2, b = 1, c = 1.
∴ b2 – 4ac = 12 – 4.2.1 = 1 – 8 = – 7
∴ x = $$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$ = $$\frac{-1 \pm \sqrt{-7}}{2.2}$$ = $$\frac{-1 \pm \sqrt{7} i}{4}$$.

3. x2 + 3x + 9 = 0
∴ a = 1, b = 3, c = 9
∴ b2 – 4ac = 32 – 4.1.9
= 9 – 36
= – 27.

4. – x2 + x – 2 = 0 or x2 – x + 2 = 0.
Here, a = 1, b = – 1, c = 2.
∴ b2 – 4ac = (- 1)2 – 4.1.2
= 1 – 8 = – 7.
∴ x = $$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$ = $$\frac{-1 \pm \sqrt{-7}}{2.1}$$ = $$\frac{-1 \pm \sqrt{7} i}{2}$$.

5. x2 + 3x + 5 = 0
∴ a = 1, b = 3, c = 5.
∴ b2 – 4ac = 9 – 4.1.5 = 9 – 20 = – 11.

6. x2 – x + 2 = 0
∴ a = 1, b = – 1, c = 2.
∴ b2 – 4ac = (- 1)2 – 4.1.2 = 1 – 8 = – 7.
∴ x = $$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$ = $$\frac{-1 \pm \sqrt{-7}}{2.1}$$ = $$\frac{-1 \pm \sqrt{7} i}{2}$$.

7. $$\sqrt{2}$$x2 + x + $$\sqrt{2}$$ = 0
∴ a = $$\sqrt{2}$$, b = 1, c = $$\sqrt{2}$$.
∴ b2 – 4ac = 12 – 4($$\sqrt{2}$$.$$\sqrt{2}$$) = 1 – 8 = – 7.
∴ $$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$ = $$\frac{-1 \pm \sqrt{-7}}{2 \cdot \sqrt{2}}$$ = $$\frac{-1 \pm \sqrt{-7} i}{2 \sqrt{2}}$$.

8. $$\sqrt{3}$$x2 – $$\sqrt{2}$$x + 3$$\sqrt{3}$$ = 0
∴ a = $$\sqrt{3}$$, b = – $$\sqrt{2}$$, c = 3$$\sqrt{3}$$.
∴ b2 – 4ac = (- $$\sqrt{2}$$)2 – 4.$$\sqrt{3}$$.3$$\sqrt{3}$$ = 2 – 36 = – 34.

9. x2 + x + $$\frac{1}{\sqrt{2}}$$ = 0, Multiplying by $$\sqrt{2}$$, we get
$$\sqrt{2x}$$ + $$\sqrt{2x}$$ + 1 = 0.
∴ a = $$\sqrt{2}$$, b = $$\sqrt{2}$$, c = 1.
∴ b2 – 4ac = ($$\sqrt{2}$$)2 – 4.$$\sqrt{2}$$.1 = 2 – 4$$\sqrt{2}$$.

10. x2 + $$\frac{x}{\sqrt{2}}$$ + 1 = 0
Multiplying by $$\sqrt{2}$$, we get
$$\sqrt{2}$$x2 + x + $$\sqrt{2}$$ = 0.
∴ a = $$\sqrt{2}$$, b = 1, c = $$\sqrt{2}$$
∴ b2 – 4ac = 12 – 4$$\sqrt{2}$$.$$\sqrt{2}$$ = 1 – 8 = – 7.
∴ x = $$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$ = $$\frac{-1 \pm \sqrt{-7}}{2 \sqrt{2}}$$ = $$\frac{-1 \pm \sqrt{7} i}{2 \sqrt{2}}$$.