Solve the following quadratic equations by factorization method

x2 + 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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