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}}).