2006 Alabama ARML TST Problems/Problem 13


Find the sum of the solutions to the equation



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:


We plug in $28-y$ for y and we get


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

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

See also

2006 Alabama ARML TST (Problems)
Preceded by:
Problem 12
Followed by:
Problem 14
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