Difference between revisions of "1986 AIME Problems/Problem 1"
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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> | + | '''<math>y(7 - y) = 12</math>''', or, by simplifying, |
− | '''< | + | '''<cmath>y^2 - 7y + 12 = (y - 3)(y - 4) = 0.</cmath>''' |
− | This means that <math>\sqrt[4]{x} = y = 3</math> or '''<math> | + | |
− | Thus the sum of the possible solutions for '''<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 = \boxed{337}</math>'''. | ||
== See also == | == See also == | ||
− | * [[ | + | {{AIME box|year=1986|before=First Question|num-a=2}} |
+ | * [[AIME Problems and Solutions]] | ||
+ | * [[American Invitational Mathematics Examination]] | ||
+ | * [[Mathematics competition resources]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:34, 22 July 2020
Problem
What is the sum of the solutions to the equation ?
Solution
Let . Then we have
, or, by simplifying,
This means that or
.
Thus the sum of the possible solutions for is
.
See also
1986 AIME (Problems • Answer Key • Resources) | ||
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 |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.