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 (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 = (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*} | ||
+ | a^4 + a^{-4} &= (a^2 + a^{-2})^2 - 2 \\ | ||
+ | &= [(a + a^{-1})^2 - 2]^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> | ||
− | 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{\ | + | 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) | ~Rbhale12 (Fundamental Logic) | ||
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~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
− | == Solution 2 | + | == Solution 3 (Detailed Explanation of Solution 2) == |
− | + | The detailed explanation of Solution 2 is as follows: | |
− | + | <cmath>\begin{alignat*}{8} | |
− | a^ | + | a+a^{-1}&=4 \\ |
− | &= \ | + | (a+a^{-1})^2&=4^2 \\ |
− | &= \boxed{\ | + | a^2+2aa^{-1}+a^{-2}&=16 \\ |
− | \end{ | + | 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 ( | + | ~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> | ||
− | + | Applying the Binomial Theorem, we raise both sides of <math>a+a^{-1}=4</math> to the fourth power: | |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\binom40a^4a^0+\binom41a^3a^{-1}+\binom42a^2a^{-2}+\binom43a^1a^{-3}+\binom44a^0a^{-4}&=256 \\ | \binom40a^4a^0+\binom41a^3a^{-1}+\binom42a^2a^{-2}+\binom43a^1a^{-3}+\binom44a^0a^{-4}&=256 \\ | ||
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\left(a^4+a^{-4}\right)+4\left(a^2+a^{-2}\right)&=250 \\ | \left(a^4+a^{-4}\right)+4\left(a^2+a^{-2}\right)&=250 \\ | ||
\left(a^4+a^{-4}\right)+4(14)&=250 \\ | \left(a^4+a^{-4}\right)+4(14)&=250 \\ | ||
− | a^4+a^{-4}&=\boxed{\ | + | a^4+a^{-4}&=\boxed{\textbf{(D)}\ 194}. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
~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> | ||
We apply the Quadratic Formula to get <math>a=2\pm\sqrt3.</math> | 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*} | <cmath>\begin{align*} | ||
− | a^4+a^{-4}&=\left(2+\sqrt{3}\right)^4 + | + | 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 | + | &=\left(2+\sqrt{3}\right)^4+\left(2-\sqrt{3}\right)^4 &&(*) \\ |
− | &=\boxed{\ | + | &=\boxed{\textbf{(D)}\ 194}. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | <b>Remarks | + | <b>Remarks about <math>\boldsymbol{(*)}</math></b> |
<ol style="margin-left: 1.5em;"> | <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>To find the fourth power of a sum/difference, we can first square that sum/difference, then square the result.</li><p> | ||
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~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*} | ||
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&1\cdot P_2-4\cdot P_1+1\cdot2 &&=0 &&\qquad\implies\qquad P_2&&=14, \\ | &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_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{\ | + | &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> | \end{alignat*}</cmath> | ||
~Albert1993 (Fundamental Logic) | ~Albert1993 (Fundamental Logic) | ||
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~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
− | == Solution | + | == Solution 7 (Answer Choices) == |
− | Note that <cmath>a^{4} + a^{-4} = | + | 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) | ~Thanosaops (Fundamental Logic) | ||
~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/MhALjut3Qmw?t=484 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == |
Latest revision as of 07:32, 4 November 2022
Contents
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.