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 (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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a^4 + a^{-4} &= \left(a^2 + a^{-2}\right)^2 - 2 \\ | a^4 + a^{-4} &= \left(a^2 + a^{-2}\right)^2 - 2 \\ | ||
&= \left[\left(a + a^{-1}\right)^2 - 2\right]^2 - 2 \\ | &= \left[\left(a + a^{-1}\right)^2 - 2\right]^2 - 2 \\ | ||
− | &= \boxed{\ | + | &= \boxed{\textbf{(D)}\ 194}. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
~Azjps (Fundamental Logic) | ~Azjps (Fundamental Logic) | ||
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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 | ||
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a^4+a^{-4}&=\left(2+\sqrt{3}\right)^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 about <math>\boldsymbol{(*)}</math></b> | <b>Remarks about <math>\boldsymbol{(*)}</math></b> | ||
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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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== Solution 6 (Answer Choices) == | == Solution 6 (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{\ | + | 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> |
~Thanosaops (Fundamental Logic) | ~Thanosaops (Fundamental Logic) | ||
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\left(a^2+a^{-2}\right)^2&=14^2 \\ | \left(a^2+a^{-2}\right)^2&=14^2 \\ | ||
a^4+2a^2a^{-2}+a^{-4}&=196 \\ | a^4+2a^2a^{-2}+a^{-4}&=196 \\ | ||
− | a^4+a^{-4}&=196-2&&=\boxed{\ | + | a^4+a^{-4}&=196-2&&=\boxed{\textbf{(D)}\ 194}. \\ |
\end{alignat*}</cmath> | \end{alignat*}</cmath> | ||
~MathFun1000 (Solution) | ~MathFun1000 (Solution) |
Revision as of 08:15, 6 September 2021
Contents
Problem
Suppose that the number satisfies the equation
. What is the value of
?
Solution 1 (Increases the Powers)
Squaring both sides of gives
from which
Squaring both sides of gives
from which
~Rbhale12 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2 (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 3 (Binomial Theorem)
Squaring both sides of gives
from which
Applying the Binomial Theorem, we raise both sides of to the fourth power:
~MRENTHUSIASM
Solution 4 (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 5 (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 6 (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)
Solution 7 (Detailed Version of Solution 1)
The algebra is as follows:
~MathFun1000 (Solution)
~MRENTHUSIASM (Minor Formatting)
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.