Difference between revisions of "2007 AMC 10A Problems/Problem 20"
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== Solution 5 (Newton's Sums) == | == Solution 5 (Newton's Sums) == | ||
− | From the first | + | From the first sentence of Solution 4, we conclude that <math>a</math> and <math>a^{-1}</math> are the roots of <math>x^2-4x+1=0.</math> Let |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
P_1&=a+a^{-1}, \ | P_1&=a+a^{-1}, \ |
Revision as of 10:07, 25 June 2021
Contents
[hide]Problem
Suppose that the number satisfies the equation . What is the value of ?
Solution 1 (Decreases the Powers)
Notice that for all real numbers we have from which We apply this result twice to get the answer: ~Azjps (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2 (Increases the Powers)
Squaring both sides of gives from which
Squaring both sides of gives from which
~Rbhale12 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 3 (Binomial Theorem)
Squaring both sides of gives from which
We raise both sides of to the fourth power, then apply the Binomial Theorem: ~MRENTHUSIASM
Solution 4 (Solves for a)
We multiply both sides of by then rearrange:
We apply the Quadratic Formula to get
Note that the roots are reciprocals of each other. Therefore, choosing either value for gives the same value for Remarks in
- To find the fourth power of a sum/difference, we can first square the sum/difference, then square the result.
- When we expand the fourth powers and combine like terms, the irrational terms will cancel.
~Azjps (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 5 (Newton's Sums)
From the first sentence of Solution 4, we conclude that and are the roots of Let From Newton's Sums, we have ~Albert1993 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 6 (Answer Choices)
Notice that We guess that is an integer, so the answer must be less than a perfect square. The only possibility is
~Thanosaops (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.