# Solve the following quadratic equations by factorization method4x2 – 12x + 25 = 0

Given 4x2 – 12x + 25 = 0

4x2 – 12x + 9 + 16 = 0

(2x)2 – 2(2x)(3) + 32 + 16 = 0

(2x – 3)2 + 16 = 0 [ (a + b)2 = a2 + 2ab + b2]

(2x – 3)2 + 16 × 1 = 0

We have i2 = –1 1 = –i2

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

(2x – 3)2 + 16(–i2) = 0

(2x – 3)2 – 16i2 = 0

(2x – 3)2 – (4i)2 = 0

Since a2 – b2 = (a + b)(a – b), we get

(2x – 3 + 4i)(2x – 3 – 4i) = 0

2x – 3 + 4i = 0 or 2x – 3 – 4i = 0

2x = 3 – 4i or 2x = 3 + 4i   Thus, the roots of the given equation are .

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