# Solve the following quadratic equation: x^2 – (2 + i) x – (1 – 7i) = 0

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x2 – (2 + i) x – (1 – 7i) = 0

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Given equation is x2 – (2 + i)x – (1 – 7i) = 0

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

a = 1, b = -(2 + i), c = -(1 – 7i)

Discriminant = b2 – 4ac

= [-(2 + i)]2 – 4 × 1 × -(1 – 7i)

= 4 + 4i + i2 + 4 – 28i

= 4 + 4i – 1 + 4 – 28i …..[∵ i = – 1]

= 7 – 24i

So, the given equation has complex roots.

These roots are given by

Let $\sqrt{7-24i}$ = a + bi, where a, b ∈ R

Squaring on both sides, we get

7 – 24i = a2 + i2 b2 + 2abi

7 – 24i = a2 – b2 + 2abi

Equating real and imaginary parts, we get

a2 – b2 = 7 and 2ab = -24