<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] Hint :
First solve for $\sqrt{z} + \sqrt{z+7} = 7 $
Next substitute $z=x^4-3x^2+5$ and solve for $x$.

Actually easy to see $\sqrt{9} + \sqrt{9+7} = 7 $.
$9$ is only solution as monotonic function has unique root (when it does).
So only need to solve for $x^4-3x^2+5=9$.
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