TITLE: Solve: $\sqrt{x + 3 + \sqrt{x + 14}} + \sqrt{x+3 - \sqrt{x + 14}} =4 $
QUESTION [6 upvotes]: I tried:
$\sqrt{x +14} = t$
$\implies \sqrt{x+3 + t} + \sqrt{x+3 - t} = 4$
But I think it is the hard way. Any ideas?

REPLY [9 votes]: The domain gives $x\geq-3$ and $(x+3)^2\geq x+14$.
Now, our equation it's
$$\left(\sqrt{x + 3 + \sqrt{x + 14}} + \sqrt{x+3 - \sqrt{x + 14}}\right)^2 =16 $$ or
$$2x+6+2\sqrt{(x+3)^2-x-14}=16$$ or
$$\sqrt{x^2+5x-5}=5-x.$$
Now, we get also $5-x\geq0$, for which we obtain
$$x^2+5x-5=x^2-10x+25$$ or
$$x=2,$$ which is valid.

REPLY [2 votes]: after squaring we get
$$x+3+\sqrt{x+14}+x+3-\sqrt{x+14}+2\sqrt{(x+3)^2-(x+4)}=16$$
simplifying we obtain
$$\sqrt{(x+3)^2-(x+14)}=5-x$$
squaring again
$$(x+3)^2-(x+14)=25+x^2-10x$$
Can you finish?