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

Latest revision as of 12:34, 22 July 2020

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

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

@@ Line 1: / Line 1: @@
 == Problem ==
-What is the sum of the solutions to the equation <math>\displaystyle \sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}</math>?
+What is the sum of the solutions to the equation <math>\sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}</math>?
 == Solution ==
 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>'''.
-Thus the sum of the possible solutions for '''<math>\displaystyle x</math>''' is '''<math>\displaystyle 4^4 + 3^4 = 337</math>''', the answer.
+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 = \boxed{337}</math>'''.
 == See also ==
-* [[1986 AIME Problems]]
+{{AIME box|year=1986|before=First Question|num-a=2}}
+* [[AIME Problems and Solutions]]
+* [[American Invitational Mathematics Examination]]
+* [[Mathematics competition resources]]
+{{MAA Notice}}

1986 AIME (Problems • Answer Key • Resources)
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All AIME Problems and Solutions