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

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Let <math>y = \sqrt[4]{x}</math>. Then we have
 
Let <math>y = \sqrt[4]{x}</math>. Then we have
 
'''<math>y(7 - y) = 12</math>''', or, by simplifying,
 
'''<math>y(7 - y) = 12</math>''', or, by simplifying,
'''<math>y^2 - 7y + 12 = (y - 3)(y - 4) = 0</math>'''.
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'''<cmath>y^2 - 7y + 12 = (y - 3)(y - 4) = 0.</cmath>'''
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This means that <math>\sqrt[4]{x} = y = 3</math> or '''<math>4</math>'''.
 
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>''', the answer.
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Thus the sum of the possible solutions for '''<math>x</math>''' is '''<math>4^4 + 3^4 = \boxed{337}</math>'''.
  
 
== See also ==
 
== See also ==
* [[1986 AIME Problems]]
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{{AIME box|year=1986|before=First Question|num-a=2}}
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* [[AIME Problems and Solutions]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
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{{MAA Notice}}

Latest revision as of 13:34, 22 July 2020

Problem

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

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 $4^4 + 3^4 = \boxed{337}$.

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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