Difference between revisions of "2007 AMC 10A Problems/Problem 20"
MRENTHUSIASM (talk | contribs) (In the next few edits, I will reformat the solutions and rearrange them by educational values. If anyone disagrees with me, please PM me--I am sure that we can figure out a solution.) |
MRENTHUSIASM (talk | contribs) |
||
Line 4: | Line 4: | ||
<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 (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*} | ||
Line 15: | Line 15: | ||
~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
− | == Solution 2 ( | + | == Solution 2 (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> | 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> | ||
Line 23: | Line 23: | ||
~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
+ | |||
+ | == Solution 3 (Binomial Theorem) == | ||
+ | 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> | ||
+ | |||
+ | We raise both sides of <math>a+a^{-1}=4</math> to the fourth power, then apply the Binomial Theorem: | ||
+ | <cmath>\begin{align*} | ||
+ | \binom40a^4a^0+\binom41a^3a^{-1}+\binom42a^2a^{-2}+\binom43a^1a^{-3}+\binom44a^0a^{-4}&=256 \ | ||
+ | a^4+4a^2+6+4a^{-2}+a^{-4}&=256 \ | ||
+ | \left(a^4+a^{-4}\right)+4\left(a^2+a^{-2}\right)&=250 \ | ||
+ | \left(a^4+a^{-4}\right)+4(14)&=250 \ | ||
+ | a^4+a^{-4}&=\boxed{\text{(D)}\ 194}. | ||
+ | \end{align*}</cmath> | ||
+ | ~MRENTHUSIASM | ||
== Solution 3 == | == Solution 3 == |
Revision as of 23:42, 24 June 2021
Contents
[hide]Problem
Suppose that the number satisfies the equation . What is the value of ?
Solution 1 (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 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 (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 3
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 4
. 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 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.