Proceed With Radical Equations $\sqrt{2x-8} + 8 = x$ I have a question about how to solve the following equation: 
$$
\sqrt{2x-8} + 8 = x
$$
It doesn't seem so difficult as we can just isolate the radical and be left with $\sqrt{2x-8} = x - 8$. further we can square both sides to get rid of the radical all together leaving us with $2x-8=x^2+64$. This however is where I get lost. I could try to factor this but it doesn't break down to pairs of binomials. I have the two answers this will generate which is $6$ and $12$, could anyone show me how to proceed from here?
Thanks!
 A: Be careful that
$$
(x-8)^2=x^2-16x+64
$$
where you forgot the $-16x$ term.
So after squaring you get
$$
2x-8=x^2-16x+64
$$
that simplifies to
$$
x^2-18x+72=0
$$
which has $6$ and $12$ as roots.
However $6$ is not a solution, because it would lead to the false equality $\sqrt{4}+8=6$.

You can get a priori conditions by setting $2x-8\ge0$ (that is $x\ge4$) for the existence of the square root, and also $x-8\ge0$ (that is $x\ge8$) before squaring, because $\sqrt{2x-8}\ge0$ by definition, so also $x-8$ must be non negative.
A: Hint:
A radical equation is equivalent to the system of an equation without radical and an inequation.
$$\sqrt A=B\iff B\ge0\enspace\text{and}\enspace A=B^2.$$
Some details:
This equation is equivalent to 
$$\begin{cases}2x-8=(x-8)^2\\
x-8\ge 0
\end{cases}\iff\begin{cases}x^2-18x+72\\
x\ge8
\end{cases} $$
The quadratic equation has an obvious interger root, $6$, hence the other root is  $72/6=12$. On ly the second root satisfies the inequality. Thus the radical equation has one solution, $12$.
A: First of all it must be $2x-8\ge 0$ for the root to exist; this gives $x\ge 4$. Then, since
$$
\sqrt{2x-8}=x-8,
$$
and since the square root is non-negative itself for all $x\ge 4$, it must also be $x-8\ge 0$, this is $x\ge 8$. Now, squaring left and right:
\begin{align}
2x-8&=x^2-16x+64\\
x^2-18x+72&=0\\
(x-6)(x-12)&=0
\end{align}
The solution $x=12$ can be accepted, while $x=6$ does not satisfy $x\ge 8$.
