<s>[INST] Solve: $\sqrt{x + 3 + \sqrt{x + 14}} + \sqrt{x+3 - \sqrt{x + 14}} =4 $ 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?
 [/INST] 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.
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