Q. 85.0( 2 Votes )
Solve the following quadratics
x2 + 2x + 2 = 0
Answer :
Given x2 + 2x + 2 = 0
⇒ x2 + 2x + 1 + 1 = 0
⇒ x2 + 2(x)(1) + 12 + 1 = 0
⇒ (x + 1)2 + 1 = 0 [∵ (a + b)2 = a2 + 2ab + b2]
We have i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
(x + 1)2 + (–i2) = 0
⇒ (x + 1)2 – i2 = 0
⇒ (x + 1)2 – (i)2 = 0
⇒ (x + 1 + i)(x + 1 – i) = 0 [∵ a2 – b2 = (a + b)(a – b)]
⇒ x + 1 + i = 0 or x + 1 – i = 0
⇒ x = –1 – i or x = –1 + i
∴ x = –1 ± i
Thus, the roots of the given equation are –1 ± i.
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