Question

How do you solve `x^2 / (x+3) - 5 / (x+3) = 0` and find any extraneous solutions?

Answer 1

`x = +-sqrt5`

Explanation:

As there are only two fractions - let's move one to the other side.
Then we can cross-multiply..

`color(blue)(x^2/(x+3) = 5/(x+3)" "larr (x+3 !=0))`

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

`x^3+3x^2= 5x+15" "larr` no fractions! Make equal to 0

`x^3+3x^2 -5x-15 =0" "larr` find factors by grouping

`x^2(x+3) -5(x+3)=0" "larr` common bracket

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

If `x+3 = 0 " "rarr x =-3` (extraneous solution `x-3 !=0)`

If `x^2 -5 = 0 " "rarr x = +-sqrt5 " "larr` the only solutions.

However if we consider the fraction right at the beginning, we could have arrived at the answer in a much quicker way!

`color(blue)(x^2/(x+3) = 5/(x+3)" "larr (x+3 !=0))`

The denominators are equal, so the numerators are equal..

`x^2 = 5`

`x = +-sqrt5`

`x+3 !=0, " so " x !=-3`