Difference between revisions of "2007 AMC 10A Problems/Problem 20"
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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}. \\ |
Revision as of 23:55, 7 October 2021
Contents
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)
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.