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

`√(-x + 6) = x - 6`

`\left(√(-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

`√(-6 + 6) = 6 - 6 \implies 0 = 0`

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

`√(-5 + 6) = 5 - 6 \implies \sqrt1 = -1 \implies 1 = -1`

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