Find all values of $x$ that satisfy the equation $x = \!\sqr...

Answer guidance

Correct answer: x=5 Solution: We first isolate the square root, so we can then square both sides to get rid of it. Subtracting 4 from both sides gives $x-4 = \!\sqrt{11-2x}$. Squaring both sides gives $x^2 - 8x + 16 = 11-2x$, or $x^2 -6x + 5=0$. Factoring gives $(x-5)(x-1)=0$, so $x=5$ or $x=1$. Because we squared the equation, we must check if our solutions are extraneous. For $x=5$, the equation reads $5 = \!\sqrt{11-10} + 4$, which is true. If $x=1$, we have $1 = \!\sqrt{11-2} + 4$, which is not true, so $x=1$ is extraneous. Therefore, our only solution is $\boxed{x=5}$.