Does a closed form solution exist for $x$?

Question

$$ \sqrt{x} + x^2 = \sqrt{2} $$ If so, how would one find it?

Answer 1

Let $u = \sqrt{x}$: $$u + u^4 = \sqrt{2}$$

There is a general formula for the quartic, though it is rather disgusting.

$$u = \frac{1}{2} \sqrt{-\frac{\sqrt[3]{\frac{1}{2} \left(9+\sqrt{3 \left(27+512 \sqrt{2}\right)}\right)}}{3^{2/3}}+\frac{4\ 2^{5/6}}{\sqrt[3]{3 \left(9+\sqrt{3 \left(27+512 \sqrt{2}\right)}\right)}}+\frac{2}{\sqrt{\frac{\sqrt[3]{\frac{1}{2} \left(9+\sqrt{3 \left(27+512 \sqrt{2}\right)}\right)}}{3^{2/3}}-\frac{4\ 2^{5/6}}{\sqrt[3]{3 \left(9+\sqrt{3 \left(27+512 \sqrt{2}\right)}\right)}}}}}-\frac{1}{2} \sqrt{\frac{\sqrt[3]{\frac{1}{2} \left(9+\sqrt{3 \left(27+512 \sqrt{2}\right)}\right)}}{3^{2/3}}-\frac{4\ 2^{5/6}}{\sqrt[3]{3 \left(9+\sqrt{3 \left(27+512 \sqrt{2}\right)}\right)}}}$$