Question -
Answer -
Given: 4x2 –12x + 25 = 0
4x2 –12x + 9 + 16 = 0
(2x)2 –2(2x)(3) + 32 + 16 = 0
(2x – 3)2 +16 = 0 [Since, (a + b)2 = a2 + 2ab + b2]
(2x – 3)2 +16 × 1 = 0
We know, i2 =–1 ⇒ 1 = –i2
By substituting 1 = –i2 inthe above equation, we get
(2x – 3)2 +16(–i2) = 0
(2x – 3)2 –16i2 = 0
(2x – 3)2 –(4i)2 = 0
[Byusing the formula, a2 – b2 = (a + b) (a – b)]
(2x – 3 + 4i) (2x – 3– 4i) = 0
2x – 3 + 4i = 0 or 2x– 3 – 4i = 0
2x = 3 – 4i or 2x = 3+ 4i
x = 3/2 – 2i or x =3/2 + 2i
∴ The roots of thegiven equation are 3/2 + 2i, 3/2 – 2i