Question

Solve for x if (2x/x-5)^2+(2x/x-5)-24=0

Answer 1

x $=\dfrac{115+\sqrt{2425}}{18}$ or $\dfrac{115-\sqrt{2425}}{18}$

Step-by-step explanation:

We have,

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

To find, the value of x = ?

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

⇒ $\dfrac{2x}{x-5}[\dfrac{2x}{x-5} +1]-24=0$

⇒ $\dfrac{2x}{x-5}[\dfrac{3x-5}{x-5} ]-24=0$

⇒ $\dfrac{2x}{x-5}[\dfrac{3x-5}{x-5} ]=24$

⇒ $2x(3x-5)=24(x-5)^2$

⇒ $x(3x-5)=12(x^2-10x+25)$

⇒ $3x^2-5x=12x^2-120x+300$

⇒ $9x^2-115x+300=0$

Here, a = 9, b = - 115 and c = 300

∴ x =  $\dfrac{-b±\sqrt{b^{2}-4ac}}{2a}$

=$\dfrac{-(-115)±\sqrt{(-115)^{2}-4(9)(300)}}{2(9)}$

$=\dfrac{115±\sqrt{13225-10800}}{18}$

$=\dfrac{115±\sqrt{2425}}{18}$

∴ x $=\dfrac{115+\sqrt{2425}}{18}$ or $\dfrac{115-\sqrt{2425}}{18}$