Question

Find the value of x in the below equation:

$\sf \bigg(\dfrac{2x}{x-5}\bigg)^2-5\bigg(\dfrac{2x}{x-5}\bigg)-24=0$

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Answer 1

Required Solution :-

$\sf\bigg(\dfrac{2x}{x-5}\bigg)^2-5\bigg(\dfrac{2x}{x-5}\bigg)-24=0$

Let's assume $\sf \dfrac{2x}{x-5}\: be \:_y$----(1.)

$\leadsto\sf y^2-5y-24=0$

Now middle term splitting::

$\longmapsto\sf y^2-8y+3y-24=0$

$\longmapsto\sf y(y-8)+3(y-8)=0$

$\longmapsto\sf (y-8)(y+3)=0$

So the value of y is 8 and -3.

Now substituting these values in equation (1.)

~For 8

$\sf \dfrac{2x}{x-5}\:=\:y$

$\longmapsto\sf \dfrac{2x}{x-5}\:=\:8$

By cross multiplication::

$\longmapsto \sf 2x=8(x-5)$

$\longmapsto \sf 2x=8x-40$

By transportation::

$\longmapsto \sf 40=8x-2x$

$\longmapsto \sf 40=6x$

$\longmapsto \sf \dfrac{40}{6}=x$

$\longmapsto \boxed{\sf \dfrac{20}{3}=x}$

~For -3

$\sf \dfrac{2x}{x-5}\:=\:y$

$\sf \dfrac{2x}{x-5}\:=\:-3$

By cross multiplication::

$\longmapsto \sf 2x=-3(x-5)$

$\longmapsto \sf 2x=-3x+15$

$\longmapsto \sf 3x+2x=15$

$\longmapsto \sf 5x=15$

$\longmapsto \sf x=\dfrac{15}{5}$

$\longmapsto \boxed{\sf x=3}$

So the required value of x are:-

$\leadsto\sf x=3,\dfrac{20}{3}$