Question -
Answer -
Given: x2 –x + 1 = 0
x2 – x+ ¼ + ¾ = 0
x2 – 2(x) (1/2) + (1/2)2 + ¾ = 0
(x – 1/2)2 +¾ = 0 [Since, (a + b)2 = a2 + 2ab + b2]
(x – 1/2)2 +¾ × 1 = 0
We know, i2 =–1 ⇒ 1 = –i2
By substituting 1 = –i2 inthe above equation, we get
(x – ½)2 +¾ (-1)2 = 0
(x – ½)2 +¾ (-i)2 = 0
(x – ½)2 –(√3i/2)2 = 0
[Byusing the formula, a2 – b2 = (a + b) (a – b)]
(x – ½ + √3i/2) (x – ½ – √3i/2) = 0
(x – ½ + √3i/2) = 0 or (x – ½ – √3i/2) = 0
x = 1/2 – √3i/2 or x = 1/2 + √3i/2
∴ The roots of thegiven equation are 1/2 + √3i/2,1/2 – √3i/2