Solving a radical equation with trinomials on both sides

Question

$$8\sqrt{a^2-4a-16}=3a^2-12a-64$$

I do know the standard procedure—square both sides, isolate square root, square again, check solutions to make sure they are real, etc. However, for a problem such as the above, how does one go about doing it?

I also know that there wouldn't be much of a problem by doing what I said; however, I'm fairly certain there's a more efficient, or at least less tedious, way of solving it.

The only clue I see is the $a^2-4a$ and $3a^2-12a$, where one would multiply to former by $3$ to get the latter, but nothing is clicking for me. If there is indeed a better way than squaring both sides, could someone point me to the right direction?

Answer 1

Hint:

Since $$3a^2-12a-64=3(a^2-4a-16)-16$$ we can make $x=\sqrt{a^2-4a-16}$ in order to solve $$8x=3x^2-16\qquad\text{restricted to }x>0$$ And then solve for $a$.

Answer 2

You need to combine the two approaches. First of all, you only need to square once. And if you substitute $p=a^2-4a$ before squaring the resulting 4th degree equation can be solved as a succession of two quadratics.