<s>[INST] Solve the inequality $x^4-3x^2+5\ge0$ 
Solve $$\sqrt{x^4-3x^2+5}+\sqrt{x^4-3x^2+12}=7.$$

$D_x:\begin{cases}x^4-3x^2+5\ge0 \\x^4-3x^2+12\ge0\end{cases}.$ We can see that $x^4-3x^2+12=(x^4-3x^2+5)+7,$ so if $x^4-3x^2+5$ is non-negative, $x^4-3x^2+12$ is also non-negative (even positive). So I am trying to solve $$x^4-3x^2+5\ge0.$$Let $x^2=y,y\ge0.$ Now we have $$y^2-3y+5\ge0; D=9-4\times5<0,a=1>0$$ so the function $f(y)=y^2-3y+5>0$ for all $y$. I don't know what to do next. The solution of the inequality is indeed $x\in(-\infty;+\infty),$ but I have the restriction $y\ge0?$
 [/INST] Denote:
$x^4-3x^2+5 = a - 7/2$
Then:
$x^4-3x^2+12 = a + 7/2$
Now you want to solve this one:
$\sqrt{a-7/2} + \sqrt{a + 7/2} = 7$
This is quite symmetrical and nice to work with.
Raise it to power $2$ and proceed, should be trivial from there.
At the end do a direct check to see if the values you found for $a$ give rise to valid roots for $x$. For example, if you get a solution for $a$  smaller than $7/2$, that obviously does not work i.e. does not give you any solutions for $x$.
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