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Solve the following quadratic equation:

(2 + i) x2 – (5 – i) x + 2(1 – i) = 0

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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

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