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Sum

**Solve the following quadratic equation: **

x^{2} + 4ix – 4 = 0

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#### Solution

Given equation is x^{2} + 4ix – 4 = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 1, b = 4i, c = – 4

Discriminant = b^{2} – 4ac

= (4i)^{2} – 4 x 1 x – 4

= 16i^{2} + 16

= – 16 + 16 ...[∵ i^{2} = – 1]

= 0

So, the given equation has equal roots.

These roots are given by

x = `(-"b" ± sqrt("b"^2 - 4"ac"))/(2"a")`

= `(-4"i" +- sqrt(0))/(2(1)`

= `(-4"i")/2`

∴ x = – 2i

∴ the roots of the given equation are – 2i and – 2i.

Concept: Solution of a Quadratic Equation in Complex Number System

Is there an error in this question or solution?

Chapter 3: Complex Numbers - Exercise 3.2 [Page 40]