# Find the maximum value of $\sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$

If $x\in\mathbb{R}$ find the maximum value of

$$\sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$$

I tried this:
Let $$y= \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$$ For maxima $\frac{dy}{dx}=0$ and $\frac{d^2y}{dx^2} < 0$. However, the equation $\frac{dy}{dx}=0$ (after simplifying and clearing the square roots) came out to be a nine degree equation which gave me a nightmare! Moreover, simplifying the derivative was also a tedious task. I found this question in my book in the chapter on theory of equation. I can't think of an algebraic solution. Please Help!
Thanks!

since $$\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$$
let $$P(x,x^2),A(3,2),B(0,1)$$ so $$|PA|-|PB|\le |AB|=\sqrt{10}$$ if and only is $A,P,B$ on a line.
• @Wintermute : The max is $\sqrt{10}$ and it occurs at one or both of the two points where the line $\overleftrightarrow{AB}$ intersects the parabola $y = x^2$. – steven gregory Aug 22 '16 at 21:57