Question

How do you solve the rational equation `x/(x-2) + (2x)/(4-x^2) = 5/(x+2)`?

Answer 1

`x=(5+-sqrt(15)i)/2`

Explanation:

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

First put all the denominatores is standard for by multiplying `(2x)/(4-x^2)`by (-1/-1)

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

`x^2-4=(x+2)(x-2)`, so we can easily put the fractions with the same denominator:

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

No we can eliminate the denominators. The equivalent of multiplying both sides of equation by `x^2-4`, but we must keep in mind that x!=+-2, because these values woul put `x^2-4=0`

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

`x^2+2x-2x=5x-10`

`x^2-5x+10=0`

Now we must use the expression for solving a second degree polynom:

`x=(5+-sqrt(5^2-4*10))/2`

`x=(5+-sqrt(-15))/2`

`x=(5+-sqrt(15)i)/2`