Solve the square root equation. $$\sqrt { 8 - x } = x - 6$$
A) There is one positive solution and one extraneous root.
B) The sum of the solutions is 11.
C) There is one negative solution and one extraneous root.
D) There are two positive solutions.
E) There are two negative solutions.

Question

Quizplus · Accepted Answer

Start by squaring both sides: $$8-x = (x-6)^2$$
This simplifies to: $$x^2 - 13x + 44 = 0$$
Factoring this quadratic gives: $$(x-4)(x-9) = 0$$
So the possible solutions are $x=4$ and $x=9$. 
But we must check for extraneous roots. Plugging in $x=4$ into the original equation gives $\sqrt{4} = -2$, which is not possible since the square root of a number is always positive. Therefore, $x=4$ is an extraneous root. Plugging in $x=9$ into the original equation gives $\sqrt{-1} = 3$, which is also not possible. Therefore, $x=9$ is an extraneous root as well. 
Thus, the only solution is the one positive root, which is $x = \boxed{5}$.

Solve the Square Root Equation 8−x=x−6\sqrt { 8 - x } = x - 68−x​=x−6

Solve the Square Root Equation $\sqrt { 8 - x } = x - 6$