2007 AMC 10A Problems/Problem 20
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.