Question

4z^-12z+25=0 solve by factorization method​

Answer 1

Answer:

See below.

Step-by-step explanation:

4z^2 - 12z  + 25 = 0

There are no real roots for the above question.

Answer 2

Given as 4x^2 – 12x + 25 = 0

4x^2 – 12x + 9 + 16 = 0

(2x)^2 – 2(2x)(3) + 32 + 16 = 0

(2x – 3)^2 + 16 = 0

[Since, (a + b)^2 = a^2 + 2ab + b^2]

(2x – 3)^2 + 16 × 1 = 0

As we know, i2 = –1 ⇒ 1 = –i2

On 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

[On using the formula, a^2 – b^2 = (a + b) (a – b)]

(2x – 3 + 4i) (2x – 3 – 4i) = 0

2x – 3 + 4i = 0 or 2x – 3 – 4i = 0

2x = 3 – 4i or 2x = 3 + 4i

x = 3/2 – 2i or x = 3/2 + 2i

∴ The roots of the given equation are 3/2 + 2i, 3/2 – 2i

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