(iii)  (2 + i)x² – (5- i)x + 2 (1 – i) = 0

applying discriminant rule,

x = (-b ±√(b² – 4ac))/2a

a = (2+i), b = -(5-i), c = 2(1-i)

since, i² = –1

substituting –1 = i²

x = (1 – i) or 4/5 – 2i/5

∴ The roots of the given equation are (1 – i), 4/5 – 2i/5

(iv) x² – (2 + i)x – (1 – 7i) = 0

applying discriminant rule,

x = (-b ±√(b² – 4ac))/2a

a = 1, b = -(2+i), c = -(1-7i)

=> 7 – 24i = 16 – 9 – 24i

7 – 24i = 16 + 9(–1) – 24i

= 16 + 9i² – 24i [∵ i² = –1]

= 4² + (3i)² – 2(4) (3i)

= (4 – 3i)² [∵ (a – b)² = a² – b² + 2ab]

7 – 24i = (4 – 3i)²

x = 3 – i or -1 + 2i

∴ The roots of the given equation are (-1 + 2i), (3 – i)