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>y(7 - y) = 12</math>''', or, by simplifying, | + | '''<math>\displaystyle y(7 - y) = 12</math>''', or, by simplifying, |
− | '''<math>y^2 - 7y + 12 = (y - 3)(y - 4) = 0</math>'''. | + | '''<math>\displaystyle y^2 - 7y + 12 = (y - 3)(y - 4) = 0</math>'''. |
− | 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>\displaystyle 4</math>'''. |
− | Thus the sum of the possible solutions for '''<math>x</math>''' is '''<math>4^4 + 3^4 = 337</math>''', the answer. | + | Thus the sum of the possible solutions for '''<math>\displaystyle x</math>''' is '''<math>\displaystyle 4^4 + 3^4 = 337</math>''', the answer. |
== See also == | == See also == | ||
* [[1986 AIME Problems]] | * [[1986 AIME Problems]] |
Revision as of 18:51, 28 October 2006
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 , the answer.