Difference between revisions of "2007 AMC 10A Problems/Problem 20"
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Suppose that the number <math>a</math> satisfies the equation <math>4 = a + a^{ - 1}</math>. What is the value of <math>a^{4} + a^{ - 4}</math>? | Suppose that the number <math>a</math> satisfies the equation <math>4 = a + a^{ - 1}</math>. What is the value of <math>a^{4} + a^{ - 4}</math>? | ||
− | <math>\ | + | <math>\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212</math> |
− | == Solution 1 ( | + | == Solution 1 (Decreases the Powers) == |
− | + | Note that for all real numbers <math>k,</math> we have <math>a^{2k} + a^{-2k} + 2 = (a^{k} + a^{-k})^2,</math> from which <cmath>a^{2k} + a^{-2k} = (a^{k} + a^{-k})^2-2.</cmath> We apply this result twice to get the answer: | |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | a^4 + a^{-4} &= | + | a^4 + a^{-4} &= (a^2 + a^{-2})^2 - 2 \ |
− | &= | + | &= [(a + a^{-1})^2 - 2]^2 - 2 \ |
− | &= \boxed{\ | + | &= \boxed{\textbf{(D)}\ 194}. |
+ | \end{align*}</cmath> | ||
~Azjps (Fundamental Logic) | ~Azjps (Fundamental Logic) | ||
~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^2+a^{-2}=14</math> gives <math>a^4+a^{-4}+2=196,</math> from which <math>a^4+a^{-4}=\boxed{\textbf{(D)}\ 194}.</math> | |
− | + | ~Rbhale12 (Fundamental Logic) | |
− | |||
− | + | ~MRENTHUSIASM (Reconstruction) | |
+ | |||
+ | == Solution 3 (Detailed Explanation of Solution 2) == | ||
+ | The detailed explanation of Solution 2 is as follows: | ||
+ | <cmath>\begin{alignat*}{8} | ||
+ | a+a^{-1}&=4 \ | ||
+ | (a+a^{-1})^2&=4^2 \ | ||
+ | a^2+2aa^{-1}+a^{-2}&=16 \ | ||
+ | a^2+a^{-2}&=16-2&&=14 \ | ||
+ | (a^2+a^{-2})^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 (Minor Formatting) | ||
+ | |||
+ | == 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> | ||
+ | |||
+ | Applying the Binomial Theorem, we raise both sides of <math>a+a^{-1}=4</math> to the fourth power: | ||
+ | <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{\textbf{(D)}\ 194}. | ||
+ | \end{align*}</cmath> | ||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == 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 apply the Quadratic Formula to get <math>a=2\pm\sqrt3.</math> | ||
+ | |||
+ | By Vieta's Formulas, note that the roots are reciprocals of each other. Therefore, both values of <math>a</math> produce the same value of <math>a^4+a^{-4}:</math> | ||
+ | <cmath>\begin{align*} | ||
+ | a^4+a^{-4}&=\left(2+\sqrt{3}\right)^4 + \left(2+\sqrt{3}\right)^{-4} \ | ||
+ | &=\left(2+\sqrt{3}\right)^4+\left(2-\sqrt{3}\right)^4 &&(*) \ | ||
+ | &=\boxed{\textbf{(D)}\ 194}. | ||
+ | \end{align*}</cmath> | ||
+ | <b>Remarks about <math>\boldsymbol{(*)}</math></b> | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>To find the fourth power of a sum/difference, we can first square that sum/difference, then square the result.</li><p> | ||
+ | <li>When we expand the fourth powers and combine like terms, the irrational terms will cancel.</li><p> | ||
+ | </ol> | ||
+ | ~Azjps (Fundamental Logic) | ||
+ | |||
+ | ~MRENTHUSIASM (Reconstruction) | ||
+ | |||
+ | == 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 | ||
+ | <cmath>\begin{align*} | ||
+ | P_1&=a+a^{-1}, \ | ||
+ | P_2&=a^2+a^{-2}, \ | ||
+ | P_3&=a^3+a^{-3}, \ | ||
+ | P_4&=a^4+a^{-4}. | ||
+ | \end{align*}</cmath> | ||
+ | By Newton's Sums, we have | ||
+ | <cmath>\begin{alignat*}{12} | ||
+ | &1\cdot P_1-4\cdot 1&&=0 &&\qquad\implies\qquad P_1&&=4, \ | ||
+ | &1\cdot P_2-4\cdot P_1+1\cdot2 &&=0 &&\qquad\implies\qquad P_2&&=14, \ | ||
+ | &1\cdot P_3-4\cdot P_2+1\cdot P_1&&=0 &&\qquad\implies\qquad P_3&&=52, \ | ||
+ | &1\cdot P_4-4\cdot P_3+1\cdot P_2&&=0 &&\qquad\implies\qquad P_4&&=\boxed{\textbf{(D)}\ 194}. | ||
+ | \end{alignat*}</cmath> | ||
+ | ~Albert1993 (Fundamental Logic) | ||
+ | |||
+ | ~MRENTHUSIASM (Reconstruction) | ||
− | == Solution | + | == Solution 7 (Answer Choices) == |
− | + | Note that <cmath>a^{4} + a^{-4} = (a^{2} + a^{-2})^{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> | |
− | + | ||
+ | ~Thanosaops (Fundamental Logic) | ||
+ | |||
+ | ~MRENTHUSIASM (Reconstruction) | ||
− | == Solution | + | == Video Solution by OmegaLearn == |
− | + | https://youtu.be/MhALjut3Qmw?t=484 | |
− | |||
− | + | ~ pi_is_3.14 | |
− | |||
== See also == | == See also == |
Latest revision as of 06:32, 4 November 2022
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.