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 ( | + | == Solution 1 (Direct) == |
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*} |
Revision as of 22:39, 24 June 2021
Contents
[hide]Problem
Suppose that the number satisfies the equation . What is the value of ?
Solution 1 (Direct)
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
Notice that . Since D is the only option 2 less than a perfect square, that is correct.
PS: Because this is a multiple choice test, this works.
Solution 3
. We apply the quadratic formula to get .
Thus (so it doesn't matter which root of we use). Using the binomial theorem we can expand this out and collect terms to get .
Solution 4
(similar to Solution 1) We know that . We can square both sides to get , so . Squaring both sides again gives , so .
Solution 5
We let and be roots of a certain quadratic. Specifically . We use Newton's Sums given the coefficients to find .
Solution 6
Let = + . Then so . Then by De Moivre's Theorem, = and solving gets 194.
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.