Difference between revisions of "2007 AMC 10A Problems/Problem 20"
MRENTHUSIASM (talk | contribs) m |
MRENTHUSIASM (talk | contribs) (Switched the order of solutions, so solutions of similar thoughts are grouped together.) |
||
Line 4: | Line 4: | ||
<math>\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212</math> | <math>\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212</math> | ||
− | == Solution 1 (Increases the Powers) == | + | == Solution 1 (Decreases the Powers) == |
+ | Note 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*} | ||
+ | a^4 + a^{-4} &= \left(a^2 + a^{-2}\right)^2 - 2 \ | ||
+ | &= \left[\left(a + a^{-1}\right)^2 - 2\right]^2 - 2 \ | ||
+ | &= \boxed{\textbf{(D)}\ 194}. | ||
+ | \end{align*}</cmath> | ||
+ | ~Azjps (Fundamental Logic) | ||
+ | |||
+ | ~MRENTHUSIASM (Reconstruction) | ||
+ | |||
+ | == 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 13: | Line 24: | ||
~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
− | == Solution 2 | + | == Solution 3 (Detailed Version of Solution 2) == |
− | + | The algebra is as follows: | |
− | + | <cmath>\begin{alignat*}{8} | |
− | a^ | + | a+a^{-1}&=4 \ |
− | &= \ | + | \left(a+a^{-1}\right)^2&=4^2 \ |
− | &= \boxed{\textbf{(D)}\ 194}. | + | a^2+2aa^{-1}+a^{-2}&=16 \ |
− | \end{ | + | a^2+a^{-2}&=16-2&&=14 \ |
− | ~ | + | \left(a^2+a^{-2}\right)^2&=14^2 \ |
+ | a^4+2a^2a^{-2}+a^{-4}&=196 \ | ||
+ | a^4+a^{-4}&=196-2&&=\boxed{\textbf{(D)}\ 194}. \ | ||
+ | \end{alignat*}</cmath> | ||
+ | ~MathFun1000 (Solution) | ||
− | ~MRENTHUSIASM ( | + | ~MRENTHUSIASM (Minor Formatting) |
− | == Solution | + | == Solution 4 (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> | 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 37: | Line 52: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == Solution | + | == Solution 5 (Solves for a) == |
We multiply both sides of <math>4=a+a^{-1}</math> by <math>a,</math> then rearrange: <cmath>a^2-4a+1=0.</cmath> | We multiply both sides of <math>4=a+a^{-1}</math> by <math>a,</math> then rearrange: <cmath>a^2-4a+1=0.</cmath> | ||
Line 57: | Line 72: | ||
~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
− | == Solution | + | == Solution 6 (Newton's Sums) == |
From the first sentence of Solution 4, we conclude that <math>a</math> and <math>a^{-1}</math> are the roots of <math>x^2-4x+1=0.</math> Let | From the first sentence of Solution 4, we conclude that <math>a</math> and <math>a^{-1}</math> are the roots of <math>x^2-4x+1=0.</math> Let | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
Line 76: | Line 91: | ||
~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
− | == Solution | + | == Solution 7 (Answer Choices) == |
Note that <cmath>a^{4} + a^{-4} = \left(a^{2} + a^{-2}\right)^{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> | Note that <cmath>a^{4} + a^{-4} = \left(a^{2} + a^{-2}\right)^{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> | ||
Line 82: | Line 97: | ||
~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
== See also == | == See also == |
Revision as of 02:17, 13 September 2021
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 Version of Solution 2)
The algebra 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.