Alternative way to calculate the sequence

Question

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 $S\approx 1.414213562\approx\sqrt 2$.

Answer 1

Rewrite using the value of $a^2 = \frac{1-a}{2\sqrt2}$

$$\frac{a+1}{\sqrt{\tfrac{(a-1)^2}{8}+a+1}+\tfrac{(a-1)}{2\sqrt{2}}} = \frac{(a+1)2\sqrt2}{\sqrt{a^2+6a+9} + a-1} = \frac{(a+1)2 \sqrt 2}{(a+3) + a-1} \\= \sqrt{2}$$

Answer 2

Hint: Use

$$4a^2+\sqrt 2 a-\sqrt 2=0 \implies a^2=\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{4}a$$

and

$$a^4 = \dfrac{2}{16}-2\dfrac{\sqrt{2}}{4}\dfrac{\sqrt{2}}{4}a+\dfrac{2}{16}a^2$$ $$= 1/8-a+1/8\left[\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{4}a\right].$$

Answer 3

$$ 4a^2 = \sqrt{2}(1-a)\Longrightarrow 8a^4 = 1-2a+a^2$$ so

$$ 8(a^4+a+1) = a^2+6a+9 = (a+3)^2$$

Answer 4

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 $(1)(2)(3)$, $$S=\frac{a+3}{2\sqrt 2}+a^2=\frac{2\sqrt 2a^2+a+3}{2\sqrt 2}=\frac{1+3}{2\sqrt 2}=\color{red}{\sqrt 2}$$