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