How sharp are your eyes? Part 3

The condition of x is $x\geq 0$

Now we have $\sqrt{x^4-x^2+4}-2x+\sqrt{x^4+20x^2+4}-5x=0$ $\Leftrightarrow (x^4-5x^2+4)\bigg(\frac{1}{\sqrt{x^4-x^2+4}+2x}+\frac{1}{\sqrt{x^4+20x^2+4}+5x}\bigg)=0$ Since $\frac{1}{\sqrt{x^4-x^2+4}+2x}+\frac{1}{\sqrt{x^4+20x^2+4}+5x} >0$ , we now have $x^4-5x^2+4=0$ $\Leftrightarrow\begin{cases} x^2=1\\x^2=4\end{cases}$ $\Leftrightarrow x=\{1;2\}$ So the sum is $3$

Since $LHS \geq 0$ , we must have $RHS \geq 0$ This implies that $x \geq 0$

Let $\sqrt{x^4-x^2+4} = b$ and $\sqrt{x^4+20x^2+4} = a$

Then we have $a + b = 7x \quad \quad - (1)$

We also have $a^2 - b^2 = (x^4+20x^2+4) - (x^4-x^2+4) = 21x^2$

Therefore $a - b = \frac{a^2 - b^2}{a + b} = 3x \quad \quad - (2)$

Now, $(1) + (2)$ gives us $2a = 10x \implies a = 5x$

$\sqrt{x^4+20x^2+4} = 5x$

$x^4+20x^2+4 = 25x^2$

$(x^2 - 4)(x^2 - 1) = 0$

Implies $x = \pm1, \pm2$ . But $x \geq 0$ . Therefore $x = 1,2$ and sum of all values is $\boxed{3}$

Substitute x^4 +19x^2 / 2 +4 = y . See the magic ! (It will simplify as if you are simplifying a simple expression)