Homework Help

# Solve the equation. `5((x+2)/(x-2))^2 - 2((x+2)/(x-2)) - 3 = 0` Hint: Use `u =...

Student

Valedictorian

• Up
• 1
• Down

Solve the equation.

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

Hint: Use `u = (x+2)/(x-2)` and solve for `x` .

Enter the solutions separated by commas.

Posted by kristenmariebieber on September 19, 2013 at 4:17 PM via web and tagged with algebra 2, math, quadratic equation

High School Teacher

(Level 3) Associate Educator

• Up
• 0
• Down

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

Substitution `u=(x+2)/(x-2)`

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

Now you use the formula for solution of quadratic equation

`x_(1,2)=(-b pm sqrt(b^2-4ac))/(2a)` where `a,b` and `c` are coefficients of the equation.

`u_(1,2)=(2pmsqrt(4+60))/(-4)=(2pm8)/(-4)`

`u_1=-5/2,` `u_2=3/2`

Now we put `u_1` back in our substitution

`-5/4=(x+2)/(x-2)`

Multiply whole equation by `4(x-2)`

`-5(x-2)=4(x+2)`

`-5x+10=4x+8`

`-9x=-2`

`x=2/9`

Now we put `u_2` back into our substitution. ` `

`3/2=(x+2)/(x-2)`

Multiply whole equation by `2(x-2)`

`3(x-2)=2(x+2)`

`3x-6=2x+4`

`x=10`

Hence your solutions are `x_1=2/9`  , `x_2=10.`

Posted by tiburtius on September 19, 2013 at 4:36 PM (Answer #1)

High School Teacher

(Level 2) Educator

• Up
• 0
• Down

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

Let

`t=(x+2)/(x-2)`  and `t!=2`  otherwise  t is not defined.

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

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

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

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

`=> t=-3/5 or t=1`

`If`

`t=1`

`(x+2)/(x-2)=1`

`x+2=x-2`

`0=4`

`=> (x+2)/(x-2)!=1`

`i.e. t=1`  not possible.

Then

`t=-3/5`

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

`5(x+2)=-3(x-2)`

`5x+10=-3x+6`

`5x+3x=6-10`

`8x=-4`

`x=-1/2`

Thus solution of above quadratic equation is x=-1/2 .

Posted by aruv on September 20, 2013 at 1:27 PM (Answer #2)