Difference between revisions of "2006 Alabama ARML TST Problems/Problem 13"

Latest revision as of 08:56, 15 December 2018

Find the sum of the solutions to the equation

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

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, the answer is $28\cdot \dfrac{4}{2}=\boxed{56}$

@@ Line 17: / Line 17: @@
 <math>\sqrt[4]{x+27}=3\Rightarrow x=54, \sqrt[4]{55-x}=1</math>
-Another solution. But we see that <math>54-26=55-27</math>. So we try to prove that if <math>y</math> is a solution, then <math>28-y</math> is a solution:
+Another solution. But we see that <math>54-26=55-27=28</math>. So we try to prove that if <math>y</math> is a solution, then <math>28-y</math> is a solution:
 <math>\sqrt[4]{y+27}+\sqrt[4]{55-y}=4</math>
@@ Line 27: / Line 27: @@
 But that just equals four! Thus, if <math>y</math> is a solution, then <math>28-y</math> is a solution.
-Since there are four roots, <math>28\cdot \dfrac{4}{2}=\boxed{56}</math>
+Since there are four roots, the answer is <math>28\cdot \dfrac{4}{2}=\boxed{56}</math>
 ==See also==
 {{ARML box|year=2006|state=Alabama|num-b=12|num-a=14}}