A fractional equation with a root to reject

Statement

Solve the fractional equation $\dfrac{x^2}{x-3} = \dfrac{9}{x-3}$, stating the acceptable solutions.

Solution

Existence condition: $x \neq 3$.

Since both sides share the same (non-zero) denominator, we may equate the numerators: $x^2 = 9 \implies x = 3 \ \vee\ x = -3.$ The value $x = 3$ must be rejected because it makes the denominator zero. Only $x = -3$ remains. Check: for $x=-3$: $\dfrac{9}{-6} = -\dfrac32$ and $\dfrac{9}{-6} = -\dfrac32$. Correct. $\boxed{x = -3}$