Exercise 5.3 Solutions Complex Numbers and Quadratic Equations – NCERT Class 11 Mathematics Chapter 5

Solve each of the following equations:

1. x2 + 3 = 0

x2 + 3 = 0

x2 = – 3

x = ± √-3

This can be rewritten as following
x = ± √3 × √-1

Replacing √-1 with Imaginary Number iota (i)

⇒ x = ± √3i

Thus value of x is ± √3i if its given that x2 + 3 = 0


2. 2x2 + x + 1 = 0

Comparing 2x2 + x + 1 = 0 with standard quadratic equation ax2 + bx + c = 0
⇒ a = 2
⇒ b = 1
⇒ c = 1

Finding out Discriminant of quadratic equation 2x2 + x + 1 = 0
D = b2 – 4ac

D = (1)2 – 4 × 2 × 1
D = 1 – 8

D = -7

Finding out solution of Quadratic Equation 2x^2 + x + 1 = 0

3. x2 + 3x + 9 = 0

Comparing x2 + 3x + 9 = 0 with standard quadratic equation ax2 + bx + c = 0
⇒ a = 1
⇒ b = 3
⇒ c = 9

Finding out Discriminant of quadratic equation x2 + 3x + 9 = 0
D = b2 – 4ac

D = (3)2 – 4 × 1 × 9
D = 9 – 36

D = -27

Finding out solution of Quadratic Equation x^2 + 3x + 9 = 0

4. – x2 + x – 2 = 0

Comparing – x2 + x – 2 = 0 with standard quadratic equation ax2 + bx + c = 0
⇒ a = – 1
⇒ b = 1
⇒ c = – 2

Finding out Discriminant of quadratic equation – x2 + x – 2 = 0
D = b2 – 4ac

D = (1)2 – 4 × (- 1) × (- 2)
D = 1 – 8

D = – 7

Finding out solution of Quadratic Equation - x^2 + x - 2 = 0

5. x2 + 3x + 5 = 0

Comparing x2 + 3x + 5 = 0 with standard quadratic equation ax2 + bx + c = 0
⇒ a = 1
⇒ b = 3
⇒ c = 5

Finding out Discriminant of quadratic equation x2 + 3x + 5 = 0
D = b2 – 4ac

D = (3)2 – 4 × 1 × 5
D = 9 – 20 = – 11

D = – 11

Finding out solution of Quadratic Equation x^2 + 3x + 5 = 0

6. x2 – x + 2 = 0

Comparing x2 – x + 2 = 0 with standard quadratic equation ax2 + bx + c = 0
⇒ a = 1
⇒ b = – 1
⇒ c = 2

Finding out Discriminant of quadratic equation x2 – x + 2 = 0
D = b2 – 4ac

D = (- 1)2 – 4 × 1 × 2
D = 1 – 8 = – 7

D = – 7

Finding out solution of Quadratic Equation x^2 - x + 2 = 0

8. √3x2 – √2x + 3√3 = 0

Comparing √3x2 – √2x + 3√3 = 0 with standard quadratic equation ax2 + bx + c = 0
⇒ a = √3
⇒ b = – √2
⇒ c = 3√3

Finding out Discriminant of quadratic equation √3x2 – √2x + 3√3 = 0
D = b2 – 4ac

D = (√3)2 – 4 × √3 × 3√3
D = 3 – 36 = – 33

D = – 33

Finding out solution of Quadratic Equation √3x^2 - √2x + 3√3 = 0

9. x2 + x + 1/√2 = 0

Simplifying Quadratic Equation x2 + x + 1/√2 = 0

√2x2 + √2x + 1 = 0

Let’s find out which values of x satisfy this Quadratic Equation

Comparing √2x2 + √2x + 1 = 0 with standard quadratic equation ax2 + bx + c = 0
⇒ a = √2
⇒ b = √2
⇒ c = 1

Finding out Discriminant of quadratic equation √2x2 + √2x + 1 = 0
D = b2 – 4ac

D = (√2)2 – 4 × √2 × 1
D = 2 – 4√2 = 2(1 – 2√2)

D = – 2(2√2 – 1)

Finding out solution of Quadratic Equation √2x^2 + √2x + 1 = 0

10. x2 + x/√2 + 1 = 0

Let’s first simplify this Quadratic Equation and then find out values of variable x.

x2 + x/√2 + 1 = 0 can be rewritten as √2x2 + x + √2 = 0

Let’s now find out values of x for which Quadratic Equation √2x2 + x + √2 = 0 is satisfied

Comparing √2x2 + x + √2 = 0 with standard quadratic equation ax2 + bx + c = 0
⇒ a = √2
⇒ b = 1
⇒ c = √2

Finding out Discriminant of quadratic equation √2x2 + √2x + 1 = 0
D = b2 – 4ac

D = (1)2 – 4 × √2 × √2
D = 1 – 4 × 2 = 1 – 8 = – 7

D = – 7

Finding out solution of Quadratic Equation √2x^2 + √2x + 1 = 0

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