Question -
Answer -
Given: x2 +2x + 5 = 0
x2 +2x + 1 + 4 = 0
x2 +2(x) (1) + 12 + 4 = 0
(x + 1)2 +4 = 0 [since, (a + b)2 = a2 + 2ab + b2]
(x + 1)2 +4 × 1 = 0
We know, i2 =–1 ⇒ 1 = –i2
By substituting 1 = –i2 inthe above equation, we get
(x + 1)2 +4(–i2) = 0
(x + 1)2 –4i2 = 0
(x + 1)2 –(2i)2 = 0
[Byusing the formula, a2 – b2 = (a + b) (a –b)]
(x + 1 + 2i)(x + 1 –2i) = 0
x + 1 + 2i = 0 or x +1 – 2i = 0
x = –1 – 2i or x = –1+ 2i
∴ The roots of thegiven equation are -1+2i, -1-2i