Question

Solve the equation. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLU X = 6 X = -x + 6 = x - 6 Identify any extraneous solution. (If there is no extraneous solution, enter NO SOLUTION.) 2​

Answer 1

Taking squares on both sides leads to

$\sqrt{-x + 6} = x - 6$

$\left(\sqrt{-x + 6}\right)^2 = (x - 6)^2$

$-x + 6 = x^2 - 12x + 36$

$x^2 - 11x + 30 = 0$

$(x - 5) (x - 6) = 0$

Solving for $x$, we get

$x - 5 = 0 \text{ or } x - 6 = 0$

or

$x = 5 \text{ or } x = 6$.

Evaluating both sides of the starting equation at these solutions, we have

$\sqrt{-6 + 6} = 6 - 6 \implies 0 = 0$

which is true, so $\boxed{x=6}$ is a valid solution. However,

$\sqrt{-5 + 6} = 5 - 6 \implies \sqrt1 = -1 \implies 1 = -1$

which is not true, so $\boxed{x=5}$ is an extraneous solution.