Question -
Answer -
Given: 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
We know, i2 =–1 ⇒ 1 = –i2
By substituting 1 = –i2 inthe above equation, we get
(x – 2)2 +3(–i2) = 0
(x – 2)2 –3i2 = 0
(x – 2)2 –(√3i)2 = 0
[Byusing 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 thegiven equation are 2 ± √3i