AoPSWiki
Do you have what it takes to be the next brilliant trader, researcher, or developer at Jane Street Capital? Find out in the Careers in Mathematics Forum.

2006 Alabama ARML TST Problems/Problem 13

From AoPSWiki

Problem

Find the sum of the solutions to the equation

\sqrt[4]{x+27}+\sqrt[4]{55-x}=4.

Solution

There are four solutions, since we have fourth roots. We try to find some nice solutions:

\sqrt[4]{x+27}=0\Rightarrow x=-27, \sqrt[4]{55-x}=\sqrt[4]{82}

Not quite, but

\sqrt[4]{x+27}=1\Rightarrow x=-26, \sqrt[4]{55-x}=\sqrt[4]{81}=3

That's a solution! Now we switch:

\sqrt[4]{x+27}=3\Rightarrow x=54, \sqrt[4]{55-x}=1

Another solution. But we see that 54-26=55-27=28. So we try to prove that if y is a solution, then 28-y is a solution:

\sqrt[4]{y+27}+\sqrt[4]{55-y}=4

We plug in 28-y for y and we get

\sqrt[4]{55-y}+\sqrt[4]{y+27}

But that just equals four! Thus, if y is a solution, then 28-y is a solution.

Since there are four roots, 28\cdot \dfrac{4}{2}=\boxed{56}

See also

2006 Alabama ARML TST (Problems)
Preceded by:
Problem 12 		Followed by:
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_Alabama_ARML_TST_Problems/Problem_13"
Trying to get to the USAMO in 2010? Our AIME Problem Series can help you get there! Click here to enroll today!
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us