Question

(2x - 3) square =4x - 6 find the value of x​

Answer 1

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• Find the value of x when $\sf {(2x - 3)}^{2} = 4x - 6$

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$\sf {(2x - 3)}^{2} = 4x - 6$

$\sf \implies 4{x}^{2} + 9 - 12x = 4x - 6$

$\sf \implies 4{x}^{2} - 16x + 15=0$

$\sf \implies 4{x}^{2} - 10x - 6x + 15=0$

$\sf \implies 2x(2x- 5) - 3(2x - 5)=0$

$\sf \implies (2x - 3)(2x - 5)=0$

Therefore, either $\sf (2x-3)=0$ or $\sf (2x-5)=0$

So,

$\sf2x - 3 = 0 \implies x = \frac{3}{2}$

$\sf2x - 5 = 0 \implies x = \frac{5}{2}$

$\star\:\:\:\sf\large\underline\blue{Answer:-}$

• The roots of the given equation are 3/2 and 5/2.

Answer 2

Answer:

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