# Find maximum of $|\sqrt{x^4-7x^2-4x+20}-\sqrt{x^4+9x^2+16}|$ without using calculus

Find maximum of $$|\sqrt{x^4-7x^2-4x+20}-\sqrt{x^4+9x^2+16}|$$ without using any calculus

I believe I have made most of the progress but am stuck on the last step

I rewrote the expression as- $$\left|\sqrt{(x^2-4)^2+(x-2)^2}-\sqrt{(x^2+4)^2+(x-0)^2}\right|$$

We notice that this is written in the format of the distance formula. In particular, this is the expression for the difference of distances from an arbitrary point $$(x^2,x)$$ and $$(4,2)$$,$$(-4,0)$$ respectively.

In this figure we have to find maximum of $$|PA-PB|$$

I could not think of how this could be maximised. Usually in questions like these $$P$$ somehow is collinear or lies on the reflection of some given point but nothing of that sort can happen here. I tried a few special points as well like $$(4,-2),(0,0)$$ etc.

Can someone help me with this?

• @TonyK yes, thanks for pointing it out Sep 23, 2020 at 17:41

The difference between the distances to each of those two points is maximised over the whole plane by any point on the straight line $$AB$$ that does not lie strictly between $$A$$ and $$B$$; and this maximum value is simply $$|AB|=2\sqrt{17}$$.
And your curve touches this line at $$A$$ when $$x=2$$; so the maximum value of the expression is $$2\sqrt{17}$$, achieved when $$x=2$$.
• I made a mistake while looking at the distance, I thought it was $7.2…$ instead of $8.2…>8$ I've edited my question so that someone in the future looking at this does not get confused; Thanks for your help Sep 23, 2020 at 17:32
By the triangle inequality, for every point $$P$$, $$||PA|-|PB||\leq |AB|$$ so the maximum is attained when $$P=A$$.