Radical maximum

Geometry Level 3

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

If the maximum value of the function above can be expressed as a \sqrt{a} , find the value of a 2 a^2 .


The answer is 100.

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Ishan Singh by Ishan Singh , 22, India

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2 solutions

Ishan Singh
Jul 23, 2014

Given : y = x 4 − 3 x 2 − 6 x + 13 − x 4 − x 2 + 1 y=\sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}

= ( x 2 − 2 ) 2 + ( x − 3 ) 2 − ( x 2 − 1 ) 2 + ( x − 0 ) 2 =\sqrt{(x^2-2)^2+(x-3)^2} - \sqrt{(x^2-1)^2+(x-0)^2}

On the Cartesian Plane if P ( x , x 2 ) , A ( 3 , 2 ) , B ( 0 , 1 ) P(x,x^2) , A(3,2) , B(0,1)
are three points, then

y = ∣ P A ∣ − ∣ P B ∣ ≤ ∣ A B ∣ = 10 y=|PA| - |PB| \le |AB|=\sqrt{10} (Using Triangle Inequality)

where the equality holds if P P , A A and B B lie on a line.

⟹ a 2 = 100 \implies a^2=\boxed{100}

37 Helpful 1 Interesting 1 Brilliant 0 Confused

Cool observation.

Nishant Sharma - 6 years, 9 months ago

Similar to mine, I used right triangles

Rindell Mabunga - 6 years, 9 months ago

Very cool problem!

Rindell Mabunga - 6 years, 9 months ago

Isn't it necessary to prove that such point P(x,x^2) exists between A and B?

Simona Vesela - 6 years ago

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No, not at all the maximum value is attained when all the three points are collinear.

Aditya Sky - 5 years, 2 months ago

done in the same way

Sagnik Dutta - 6 years, 9 months ago
Emmanuel Torres
Feb 9, 2017

I plugged in x = -1. And i got √17 - √1, which is about 3.12. 3.12^2 gives you about 9.75. Which is decent, but not best. So I assumed the best was 10. So, 10^2 gives you 100.

1 Helpful 0 Interesting 0 Brilliant 1 Confused

Lucky...but not an explain

Isaac YIU Math Studio - 1 year, 10 months ago

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