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Difference between revisions of "2007 AMC 10A Problems/Problem 20"

(Switched the order of solutions, so solutions of similar thoughts are grouped together.)
(→‎Solution 3 (Detailed Explanation of Solution 2))
 
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== Solution 1 (Decreases 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:
+
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} &= \left(a^2 + a^{-2}\right)^2 - 2 \
+
a^4 + a^{-4} &= (a^2 + a^{-2})^2 - 2 \
−
&= \left[\left(a + a^{-1}\right)^2 - 2\right]^2 - 2 \
+
&= [(a + a^{-1})^2 - 2]^2 - 2 \
 
&= \boxed{\textbf{(D)}\ 194}.
 
&= \boxed{\textbf{(D)}\ 194}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
~Azjps (Fundamental Logic)
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~Azjps ~MRENTHUSIASM
−
 
 
−
~MRENTHUSIASM (Reconstruction)
 
  
 
== Solution 2 (Increases the Powers) ==
 
== Solution 2 (Increases the Powers) ==
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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>
 
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)
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~Rbhale12 ~MRENTHUSIASM
  
−
~MRENTHUSIASM (Reconstruction)
+
== Solution 3 (Detailed Explanation of Solution 2) ==
−
 
+
The detailed explanation of Solution 2 is as follows:
−
== Solution 3 (Detailed Version of Solution 2) ==
 
−
The algebra is as follows:
 
 
<cmath>\begin{alignat*}{8}
 
<cmath>\begin{alignat*}{8}
 
a+a^{-1}&=4 \
 
a+a^{-1}&=4 \
−
\left(a+a^{-1}\right)^2&=4^2 \
+
(a+a^{-1})^2&=4^2 \
 
a^2+2aa^{-1}+a^{-2}&=16 \
 
a^2+2aa^{-1}+a^{-2}&=16 \
 
a^2+a^{-2}&=16-2&&=14 \
 
a^2+a^{-2}&=16-2&&=14 \
−
\left(a^2+a^{-2}\right)^2&=14^2 \
+
(a^2+a^{-2})^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{\textbf{(D)}\ 194}. \
 
a^4+a^{-4}&=196-2&&=\boxed{\textbf{(D)}\ 194}. \
 
\end{alignat*}</cmath>
 
\end{alignat*}</cmath>
−
~MathFun1000 (Solution)
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~MathFun1000 ~MRENTHUSIASM
−
 
 
−
~MRENTHUSIASM (Minor Formatting)
 
  
 
== Solution 4 (Binomial Theorem) ==
 
== Solution 4 (Binomial Theorem) ==
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<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
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{\textbf{(D)}\ 194}.
 
&=\boxed{\textbf{(D)}\ 194}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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   <li>When we expand the fourth powers and combine like terms, the irrational terms will cancel.</li><p>
 
   <li>When we expand the fourth powers and combine like terms, the irrational terms will cancel.</li><p>
 
</ol>
 
</ol>
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~Azjps (Fundamental Logic)
+
~Azjps ~MRENTHUSIASM
−
 
 
−
~MRENTHUSIASM (Reconstruction)
 
  
 
== Solution 6 (Newton's Sums) ==
 
== Solution 6 (Newton's Sums) ==
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&1\cdot P_4-4\cdot P_3+1\cdot P_2&&=0 &&\qquad\implies\qquad P_4&&=\boxed{\textbf{(D)}\ 194}.
 
&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)
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~Albert1993 ~MRENTHUSIASM
  
−
~MRENTHUSIASM (Reconstruction)
+
== 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>
  
−
== Solution 7 (Answer Choices) ==
+
~Thanosaops ~MRENTHUSIASM
−
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)
+
== Video Solution by OmegaLearn ==
 +
https://youtu.be/MhALjut3Qmw?t=484
  
−
~MRENTHUSIASM (Reconstruction)
+
~ pi_is_3.14
  
 
== See also ==
 
== See also ==

Latest revision as of 11:41, 7 March 2025

Problem

Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$. What is the value of $a^{4} + a^{ - 4}$?

$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$

Solution 1 (Decreases the Powers)

Note that for all real numbers $k,$ we have $a^{2k} + a^{-2k} + 2 = (a^{k} + a^{-k})^2,$ from which \[a^{2k} + a^{-2k} = (a^{k} + a^{-k})^2-2.\] We apply this result twice to get the answer: \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*} ~Azjps ~MRENTHUSIASM

Solution 2 (Increases the Powers)

Squaring both sides of $a+a^{-1}=4$ gives $a^2+a^{-2}+2=16,$ from which $a^2+a^{-2}=14.$

Squaring both sides of $a^2+a^{-2}=14$ gives $a^4+a^{-4}+2=196,$ from which $a^4+a^{-4}=\boxed{\textbf{(D)}\ 194}.$

~Rbhale12 ~MRENTHUSIASM

Solution 3 (Detailed Explanation of Solution 2)

The detailed explanation of Solution 2 is as follows: \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*} ~MathFun1000 ~MRENTHUSIASM

Solution 4 (Binomial Theorem)

Squaring both sides of $a+a^{-1}=4$ gives $a^2+a^{-2}+2=16,$ from which $a^2+a^{-2}=14.$

Applying the Binomial Theorem, we raise both sides of $a+a^{-1}=4$ to the fourth power: \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*} ~MRENTHUSIASM

Solution 5 (Solves for a)

We multiply both sides of $4=a+a^{-1}$ by $a,$ then rearrange: \[a^2-4a+1=0.\]

We apply the Quadratic Formula to get $a=2\pm\sqrt3.$

By Vieta's Formulas, note that the roots are reciprocals of each other. Therefore, both values of $a$ produce the same value of $a^4+a^{-4}:$ \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*} Remarks about $\boldsymbol{(*)}$

  1. To find the fourth power of a sum/difference, we can first square that sum/difference, then square the result.

  2. When we expand the fourth powers and combine like terms, the irrational terms will cancel.

~Azjps ~MRENTHUSIASM

Solution 6 (Newton's Sums)

From the first sentence of Solution 4, we conclude that $a$ and $a^{-1}$ are the roots of $x^2-4x+1=0.$ Let \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*} By Newton's Sums, we have \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*} ~Albert1993 ~MRENTHUSIASM

Solution 7 (Answer Choices)

Note that \[a^{4} + a^{-4} = (a^{2} + a^{-2})^{2} - 2.\] We guess that $a^{2} + a^{-2}$ is an integer, so the answer must be $2$ less than a perfect square. The only possibility is $\boxed{\textbf{(D)}\ 194}.$

~Thanosaops ~MRENTHUSIASM

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

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