Alternative way to calculate the sequence Given $a>0$ that satisfies $4a^2+\sqrt 2 a-\sqrt 2=0$.
Calculate $S=\frac{a+1}{\sqrt{a^4+a+1}-a^2}$.
Attempt:
There is only one number $a>0$ that satisfies $4a^2+\sqrt 2 a-\sqrt 2=0$, that is
$a=\frac{-\sqrt{2}+\sqrt{\Delta }}{2\times 4}=\frac{-\sqrt{2}+\sqrt{2+16\sqrt{2}}}{8}$
However obviously if you replace it directly to calculate $S$, it would be extremely time-consuming.
Is there another way? I think the first equation can be changed (without needing to solve it) to calculate $S$, as when I use brute force (with calculator), I found out that $S\approx 1.414213562\approx\sqrt 2$.

Hint: Use
$$4a^2+\sqrt 2 a-\sqrt 2=0 \implies a^2=\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{4}a$$
and 
$$a^4 = \dfrac{2}{16}-2\dfrac{\sqrt{2}}{4}\dfrac{\sqrt{2}}{4}a+\dfrac{2}{16}a^2$$
$$= 1/8-a+1/8\left[\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{4}a\right].$$
