# Solve the equation. 5(x+2/x-2)^2 - 2(x+2/x-2) - 3 = 0 Enter the solutions, separated by commas.

*print*Print*list*Cite

Student Comments

aruv | Student

`5(x+2/x-2)^2-2(x+2/x-2)-3=0`

Let

`x+2/x-2=y`

Then

`5(y)^2-2y-3=0`

`5y^2-2y-3=0`

`5y^2-5y+3y-3=0`

`(y-1)(5y+3)=0`

Either

y-1=0

y=1

or

5y+3=0

5y=-3

y=-3/5

Thus

`x+2/x-2=1`

`x+2/x=3`

`x^2+2=3x`

`x^2-3x+2=0`

`x^2-2x-x+2=0`

`(x-2)(x-1)=0`

`either x-2=0 or x-1=0`

`x=2 or x=1`

and

`x+2/x-2=-3/5`

`x+2/x=-3/5+2`

`(x^2+2)/x=7/10`

`10x^2+20=7x`

`10x^2-7x+20=0`

`x=(7+-sqrt(49-800))/20`

`x=(7+-sqrt(-751))/20`

`x_{1,2}=(7+-sqrt(751)i)/20`

`` Thus roots of the equation are

`x=1,2 ,(7+sqrt(751)i)/20,(7-sqrt(751)i)/20`