Difference between revisions of "1986 AIME Problems/Problem 1"

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(Solution)
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== Solution ==
 
== Solution ==
 
Let <math>y = \sqrt[4]{x}</math>. Then we have
 
Let <math>y = \sqrt[4]{x}</math>. Then we have
'''<math>\displaystyle y(7 - y) = 12</math>''', or, by simplifying,
+
'''<math>y(7 - y) = 12</math>''', or, by simplifying,
'''<math>\displaystyle y^2 - 7y + 12 = (y - 3)(y - 4) = 0</math>'''.
+
'''<cmath>y^2 - 7y + 12 = (y - 3)(y - 4) = 0.</cmath>'''
This means that <math>\sqrt[4]{x} = y = 3</math> or '''<math>\displaystyle 4</math>'''.
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Thus the sum of the possible solutions for '''<math>\displaystyle x</math>''' is '''<math>\displaystyle 4^4 + 3^4 = 337</math>''', the answer.
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This means that <math>\sqrt[4]{x} = y = 3</math> or '''<math>4</math>'''.
 +
 
 +
Thus the sum of the possible solutions for '''<math>x</math>''' is '''<math>4^4 + 3^4 = 337</math>'''.
  
 
== See also ==
 
== See also ==

Revision as of 14:43, 7 July 2008

Problem

What is the sum of the solutions to the equation $\displaystyle \sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}$?

Solution

Let $y = \sqrt[4]{x}$. Then we have $y(7 - y) = 12$, or, by simplifying, \[y^2 - 7y + 12 = (y - 3)(y - 4) = 0.\]

This means that $\sqrt[4]{x} = y = 3$ or $4$.

Thus the sum of the possible solutions for $x$ is $4^4 + 3^4 = 337$.

See also

1986 AIME (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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