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 $S\approx 1.414213562\approx\sqrt 2$.
 A: Rewrite using the value of $a^2 = \frac{1-a}{2\sqrt2}$
$$\frac{a+1}{\sqrt{\tfrac{(a-1)^2}{8}+a+1}+\tfrac{(a-1)}{2\sqrt{2}}} = \frac{(a+1)2\sqrt2}{\sqrt{a^2+6a+9}  + a-1} = \frac{(a+1)2 \sqrt 2}{(a+3) + a-1} \\= \sqrt{2}$$
A: First, we have
$$\begin{align}S&=\frac{a+1}{\sqrt{a^4+a+1}-a^2}\\\\&=\frac{a+1}{\sqrt{a^4+a+1}-a^2}\cdot\frac{\sqrt{a^4+a+1}+a^2}{\sqrt{a^4+a+1}+a^2}\\\\&=\frac{(a+1)(\sqrt{a^4+a+1}+a^2)}{a+1}\\\\&=\sqrt{a^4+a+1}+a^2\tag1\end{align}$$
We have
$$4a^2+\sqrt 2 a-\sqrt 2=0\implies (4a^2)^2=(\sqrt 2-\sqrt 2\ a)^2\implies a^4=\frac{a^2-2a+1}{8}$$
from which 
$$a^4+a+1=\frac{a^2-2a+1}{8}+a+1=\frac{(a+3)^2}{8}\tag2$$
follows.
Also, we have
$\sqrt 2\ (4a^2+\sqrt 2 a-\sqrt 2)=0\implies 2\sqrt 2\ a^2+a=1\tag3$
From $(1)(2)(3)$,
$$S=\frac{a+3}{2\sqrt 2}+a^2=\frac{2\sqrt 2a^2+a+3}{2\sqrt 2}=\frac{1+3}{2\sqrt 2}=\color{red}{\sqrt 2}$$
A: Hint: Use
$$4a^2+\sqrt 2 a-\sqrt 2=0 \implies a^2=\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{4}a$$
and 
$$a^4 = \dfrac{2}{16}-2\dfrac{\sqrt{2}}{4}\dfrac{\sqrt{2}}{4}a+\dfrac{2}{16}a^2$$
$$= 1/8-a+1/8\left[\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{4}a\right].$$
A: $$ 4a^2 = \sqrt{2}(1-a)\Longrightarrow 8a^4 = 1-2a+a^2$$ so 
$$  8(a^4+a+1) = a^2+6a+9 = (a+3)^2$$
