Difference between revisions of "2007 AMC 10A Problems/Problem 20"
MRENTHUSIASM (talk | contribs) m (→Solution 1 (Decreases the Powers)) |
Pi is 3.14 (talk | contribs) (→Solution 7 (Answer Choices)) |
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<cmath>\begin{alignat*}{8} | <cmath>\begin{alignat*}{8} | ||
a+a^{-1}&=4 \ | a+a^{-1}&=4 \ | ||
− | + | (a+a^{-1})^2&=4^2 \ | |
a^2+2aa^{-1}+a^{-2}&=16 \ | a^2+2aa^{-1}+a^{-2}&=16 \ | ||
a^2+a^{-2}&=16-2&&=14 \ | a^2+a^{-2}&=16-2&&=14 \ | ||
− | + | (a^2+a^{-2})^2&=14^2 \ | |
a^4+2a^2a^{-2}+a^{-4}&=196 \ | a^4+2a^2a^{-2}+a^{-4}&=196 \ | ||
a^4+a^{-4}&=196-2&&=\boxed{\textbf{(D)}\ 194}. \ | a^4+a^{-4}&=196-2&&=\boxed{\textbf{(D)}\ 194}. \ | ||
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
a^4+a^{-4}&=\left(2+\sqrt{3}\right)^4 + \left(2+\sqrt{3}\right)^{-4} \ | a^4+a^{-4}&=\left(2+\sqrt{3}\right)^4 + \left(2+\sqrt{3}\right)^{-4} \ | ||
− | &=\left(2+\sqrt{3}\right)^4+\left(2-\sqrt{3}\right)^4 | + | &=\left(2+\sqrt{3}\right)^4+\left(2-\sqrt{3}\right)^4 &&(*) \ |
&=\boxed{\textbf{(D)}\ 194}. | &=\boxed{\textbf{(D)}\ 194}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
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== Solution 7 (Answer Choices) == | == Solution 7 (Answer Choices) == | ||
− | Note that <cmath>a^{4} + a^{-4} = | + | Note that <cmath>a^{4} + a^{-4} = (a^{2} + a^{-2})^{2} - 2.</cmath> We guess that <math>a^{2} + a^{-2}</math> is an integer, so the answer must be <math>2</math> less than a perfect square. The only possibility is <math>\boxed{\textbf{(D)}\ 194}.</math> |
~Thanosaops (Fundamental Logic) | ~Thanosaops (Fundamental Logic) | ||
~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/MhALjut3Qmw?t=484 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == |
Latest revision as of 06:32, 4 November 2022
Contents
[hide]Problem
Suppose that the number satisfies the equation . What is the value of ?
Solution 1 (Decreases the Powers)
Note 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 (Detailed Explanation of Solution 2)
The detailed explanation of Solution 2 is as follows: ~MathFun1000 (Solution)
~MRENTHUSIASM (Minor Formatting)
Solution 4 (Binomial Theorem)
Squaring both sides of gives from which
Applying the Binomial Theorem, we raise both sides of to the fourth power: ~MRENTHUSIASM
Solution 5 (Solves for a)
We multiply both sides of by then rearrange:
We apply the Quadratic Formula to get
By Vieta's Formulas, note that the roots are reciprocals of each other. Therefore, both values of produce the same value of Remarks about
- To find the fourth power of a sum/difference, we can first square that 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 6 (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 7 (Answer Choices)
Note 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)
Video Solution by OmegaLearn
https://youtu.be/MhALjut3Qmw?t=484
~ pi_is_3.14
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.