TITLE: Algebra square roots addition problem $\sqrt {x+15} + \sqrt x = 15$
QUESTION [5 upvotes]: Saw this algebra problem in an ad, and don't immediately remember how to solve it:
$\sqrt {x+15} + \sqrt x = 15$
I at least remember that you can't simply square everything, i.e., if a + b = c, it does not necessarily follow that $\ a^2 + b^2 = c^2$.
Of course, it does follow that $(\sqrt {x+15} + \sqrt x)^2 = 15^2 = 225$, but not sure if that's the right direction to go in, or where to go from there if it is. I don't want anyone to solve it for me, but please throw me a hint if you'd be so kind - thanks!

REPLY [2 votes]: The key here is that the two radicals are of the form $\sqrt{x+a}$. If you have
$$
\sqrt{x+a}+\sqrt{x+b}=k \tag{1}
$$
with $a>b$ and $k>0$, then you can multiply both sides by $\sqrt{x+a}-\sqrt{x+b}$, getting
$$
(x+a)-(x+b)=k(\sqrt{x+a}-\sqrt{x+b})
$$
so
$$
\sqrt{x+a}-\sqrt{x+b}=\frac{a-b}{k}\tag{2}
$$
Now look at equations (1) and (2)…

This would also work for equations of the form
$$
\sqrt{mx+a}+\sqrt{mx+b}=k
$$
or
$$
\sqrt{mx+a}-\sqrt{mx+b}=k
$$
but not if the coefficients of $x$ are different in the two radicals. In that case, squaring (and hoping for the best) is the way to go.

You could also square: $(\sqrt{x+15}+\sqrt{x})^2=15^2$ so $2x+15+2\sqrt{x(x+15)}=225$ and finally
$$
\sqrt{x(x+15)}=105-x
$$
Squaring again,
$$
x^2+15x=11025-210x+x^2
$$
or
$$
225x=11025
$$
Not really a big deal, but the above methods has smaller figures.