Question

How do you solve rational equations `((x^2)/(x^2-4)) = ((x)/(x 2)) - ((2x)/(2-x))`?

Answer 1

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

`implies x^2/((x-2)(x+2))=x/x^2+(2x)/(x-2)`

Here we can see that `L.C.M` is `x(x-2)(x+2)`
Multiply both sides by `L.C.M` i.e. by `x(x-2)(x+2)`

`implies (x(x-2)(x+2)x^2)/((x-2)(x+2))=(x(x-2)(x+2)x)/x^2+(x(x-2)(x+2)2x)/(x-2)`

`implies x^3=(x-2)(x+2)+2x^2(x+2)`
`implies x^3=x^2-4+2x^3+4x^2`
`implies x^3 +5x^2-4=0`
`implies x^3+x^2+4x^2+4x-4x-4=0`
`implies x^2(x+1)+4x(x+1)-4(x+1)=0`
`implies (x+1)(x^2+4x-4)=0`
`implies` either `x+1=0 or x^2+4x-4=0`

`implies x=-1 or x=(-4+-sqrt(4^2-4*1(-4)))/(2*1)`

`implies x=-1 or x=(-4+-sqrt(16+16))/2`

`implies x=-1 or x=(-4+-4sqrt(1+1))/2`

`implies x=-1 or x=(2(-2+-2sqrt(1+1)))/2`

`implies x=-1 or x=-2+-2sqrt(2)`

Hence `x` can have three values which are `x=-1, -2+2sqrt2, -2-2sqrt2`