Difference between revisions of "2007 AMC 10A Problems/Problem 20"
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<math>\text{(A)}\ 164 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 192 \qquad \text{(D)}\ 194 \qquad \text{(E)}\ 212</math> | <math>\text{(A)}\ 164 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 192 \qquad \text{(D)}\ 194 \qquad \text{(E)}\ 212</math> | ||
− | == Solution 1 (Decreases the Powers) == | + | == Solution 1 (Increases the Powers) == |
+ | Squaring both sides of <math>a+a^{-1}=4</math> gives <math>a^2+a^{-2}+2=16,</math> from which <math>a^2+a^{-2}=14.</math> | ||
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+ | Squaring both sides of <math>a^2+a^{-2}=14</math> gives <math>a^4+a^{-4}+2=196,</math> from which <math>a^4+a^{-4}=\boxed{\text{(D)}\ 194}.</math> | ||
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+ | ~Rbhale12 (Fundamental Logic) | ||
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+ | ~MRENTHUSIASM (Reconstruction) | ||
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+ | == Solution 2 (Decreases the Powers) == | ||
Notice that for all real numbers <math>k,</math> we have <math>a^{2k} + a^{-2k} + 2 = \left(a^{k} + a^{-k}\right)^2,</math> from which <cmath>a^{2k} + a^{-2k} = \left(a^{k} + a^{-k}\right)^2-2.</cmath> We apply this result twice to get the answer: | Notice that for all real numbers <math>k,</math> we have <math>a^{2k} + a^{-2k} + 2 = \left(a^{k} + a^{-k}\right)^2,</math> from which <cmath>a^{2k} + a^{-2k} = \left(a^{k} + a^{-k}\right)^2-2.</cmath> We apply this result twice to get the answer: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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\end{align*}</cmath> | \end{align*}</cmath> | ||
~Azjps (Fundamental Logic) | ~Azjps (Fundamental Logic) | ||
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~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) |
Revision as of 12:37, 25 June 2021
Contents
[hide]Problem
Suppose that the number satisfies the equation . What is the value of ?
Solution 1 (Increases the Powers)
Squaring both sides of gives from which
Squaring both sides of gives from which
~Rbhale12 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2 (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 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 By 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.