# 2006 Alabama ARML TST Problems/Problem 13

## Problem

Find the sum of the solutions to the equation $\sqrt{x+27}+\sqrt{55-x}=4.$

## Solution

There are four solutions, since we have fourth roots. We try to find some nice solutions: $\sqrt{x+27}=0\Rightarrow x=-27, \sqrt{55-x}=\sqrt{82}$

Not quite, but $\sqrt{x+27}=1\Rightarrow x=-26, \sqrt{55-x}=\sqrt{81}=3$

That's a solution! Now we switch: $\sqrt{x+27}=3\Rightarrow x=54, \sqrt{55-x}=1$

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: $\sqrt{y+27}+\sqrt{55-y}=4$

We plug in $28-y$ for y and we get $\sqrt{55-y}+\sqrt{y+27}$

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 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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