Question

solve for
x:
(2x/x-5)^2+5(2x/x-5)-24=0

where x is not equal to 5

it's urgent please send as fast as possible.​

Answer 1

Given:-

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

To find:-

• The value of x

Answer:-

Given that,

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

Let $\sf \dfrac{2x}{x - 5}$ be $\sf a.$

$\longrightarrow \sf {a}^{2} + 5a - 24 = 0$

On splitting the middle term,

$\longrightarrow \sf {a}^{2} + 8a - 3a - 24 = 0$

$\longrightarrow \sf a(a + 8) - 3(a + 8) = 0$

$\longrightarrow \sf (a + 8)(a - 3) = 0$

So,

If (a + 8) = 0,

$\sf \: a = - 8$

Substituting the value of $\sf a$ which we assumed earlier,

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

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

$\sf \longrightarrow 2x = - 8x + 40$

$\sf \longrightarrow 10x = 40$

$\longrightarrow \underline{ \underline{ \sf{ \green{ x_{1} = 4}}}}$

Or if (a - 3) = 0,

$\sf \: a = 3$

Substituting the value of $\sf a$ which we assumed earlier,

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

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

$\sf \longrightarrow 2x = 3x - 15$

$\longrightarrow \underline{ \underline{ \sf{ \green{ x_{2} = 15}}}}$

Answer 2

let 2x÷x-5 be y. =y square+5y-24=0

(y+8) (y-3) =0

y=3 , -8

putting. y=3

2x÷x-5 =3

2x=3x-15

x=15

or. 2x÷x-5=-8

2x=-8x+40

10x=40

x=4

Hence , x=15 , 4