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?

  • $\begingroup$ it would get easier if you take $b=a-2$ $\endgroup$
    – mint
    Commented Jan 25, 2016 at 19:11

2 Answers 2



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 $a$.


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .