Given as x2 – 4x + 7 = 0
x2 – 4x + 4 + 3 = 0
x2 – 2(x) (2) + 22 + 3 = 0
(x – 2)2 + 3 = 0 [Since, (a – b)2 = a2 – 2ab + b2]
(x – 2)2 + 3 × 1 = 0
As we know, i2 = –1 ⇒ 1 = –i2
On substituting 1 = –i2 in the above equation, we get
(x – 2)2 + 3(–i2) = 0
(x – 2)2 – 3i2 = 0
(x – 2)2 – (√3i)2 = 0
[On using the formula, a2 – b2 = (a + b) (a – b)]
(x – 2 + √3i) (x – 2 – √3i) = 0
(x – 2 + √3i) = 0 or (x – 2 – √3i) = 0
x = 2 – √3i or x = 2 + √3i
x = 2 ± √3i
∴ The roots of the given equation are 2 ± √3i
