Question

Solve the following quadratic equation:

(2 + i) x² – (5 – i) x + 2(1 – i) = 0

Answer 1

Given equation is (2 + i) x² – (5 – i) x + 2(1 – i) = 0

Comparing with ax² + bx + c = 0, we get

a = 2 + i, b = -(5 – i), c = 2(1 – i)

Discriminant = b² – 4ac

= [-(5 – i)]² – 4 × (2 + i) × 2(1 – i)

= 25 – 10i + i² – 8(2 + i) (1 – i)

= 25 – 10i + i² – 8(2 – 2i + i – i² )

= 25 – 10i – 1 – 8(2 – i + 1) …..[∵ i² = -1]

= 25 – 10i – 1 – 16 + 8i – 8

= -2i

So, the given equation has complex roots.

These roots are given by

Let \(\sqrt{-2i}\) = a + bi, where a, b ∈ R

Squaring on both sides, we get

-2i = a² + b² i² + 2abi 

-2i = a² – b² + 2abi

Equating real and imaginary parts, we get

a² – b² = 0 and 2ab = -2

a² – b² = 0 and b = -1/a