Proceed With Radical Equations $\sqrt{2x-8} + 8 = x$

Question

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!

Answer 1

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 $\sqrt{4}+8=6$.

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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.

Answer 2

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, $12$.