# Solve the following quadraticsx2 + 2x + 2 = 0

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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