# Solve the following quadratic equations by factorization methodx2 + 2x + 5 = 0

Given x2 + 2x + 5 = 0

x2 + 2x + 1 + 4 = 0

x2 + 2(x)(1) + 12 + 4 = 0

(x + 1)2 + 4 = 0 [ (a + b)2 = a2 + 2ab + b2]

(x + 1)2 + 4 × 1 = 0

We have i2 = –1 1 = –i2

By substituting 1 = –i2 in the above equation, we get

(x + 1)2 + 4(–i2) = 0

(x + 1)2 – 4i2 = 0

(x + 1)2 – (2i)2 = 0

(x + 1 + 2i)(x + 1 – 2i) = 0 [ a2 – b2 = (a + b)(a – b)]

x + 1 + 2i = 0 or x + 1 – 2i = 0

x = –1 – 2i or x = –1 + 2i

x = –1 ± 2i

Thus, the roots of the given equation are –1 ± 2i.

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