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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>?
  
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<math>\text{(A)}\ 164 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 192 \qquad \text{(D)}\ 194 \qquad \text{(E)}\ 212</math>
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<math>\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212</math>
  
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__NOTOC__
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== Solution 1 (Decreases the Powers) ==
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== Solutions ==
+
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:
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=== Solution 1 ===
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<cmath>\begin{align*}
−
Notice that <math>(a^{k} + a^{-k})^2 = a^{2k} + a^{-2k} + 2</math>. Thus <math>a^4 + a^{-4} = (a^2 + a^{-2})^2 - 2 = [(a + a^{-1})^2 - 2]^2 - 2 = 194\ \mathrm{(D)}</math>.
+
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)
  
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=== Solution 2 ===
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~MRENTHUSIASM (Reconstruction)
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Notice that <math>(a^{4} + a^{-4}) = (a^{2} + a^{-2})^{2} - 2</math>. Since D is the only option 2 less than a perfect square, that is correct.
 
  
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PS: Because this is a multiple choice test, this works.
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== Solution 2 (Increases the Powers) ==
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LOL
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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>
  
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=== Solution 3 ===
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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>
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<math>4a = a^2 + 1</math>. We apply the [[quadratic formula]] to get <math>a = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}</math>.
 
  
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Thus <math>a^4 + a^{-4} = (2+\sqrt{3})^4 + \frac{1}{(2+\sqrt{3})^4} = (2+\sqrt{3})^4 + (2-\sqrt{3})^4</math> (so it doesn't matter which root of <math>a</math> we use). Using the [[binomial theorem]] we can expand this out and collect like terms to get <math>194</math>.
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~Rbhale12 (Fundamental Logic)
  
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=== Solution 4 ===
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~MRENTHUSIASM (Reconstruction)
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(similar to Solution 1)
 
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We know that <math>a+\frac{1}{a}=4</math>. We can square both sides to get <math>a^2+\frac{1}{a^2}+2=16</math>, so <math>a^2+\frac{1}{a^2}=14</math>. Squaring both sides again gives <math>a^4+\frac{1}{a^4}+2=14^2=196</math>, so <math>a^4+\frac{1}{a^4}=\boxed{194}</math>.
 
  
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=== Solution 5 ===
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== Solution 3 (Detailed Explanation of Solution 2) ==
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We let <math>a</math> and <math>1/a</math> be roots of a certain quadratic. Specifically <math>x^2-4x+1=0</math>. We use [[Newton's Sums]] given the coefficients to find <math>S_4</math>.
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The detailed explanation of Solution 2 is as follows:
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<math>S_4=\boxed{194}</math>
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<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)
  
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=== Solution 6 ===
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~MRENTHUSIASM (Minor Formatting)
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Let <math>a</math> = <math>\cos(x)</math> + <math>i\sin(x)</math>. Then <math>a + a^{-1} = 2\cos(x)</math> so <math>\cos(x) = 2</math>. Then by [[De Moivre's Theorem]], <math>a^4 + a^{-4}</math> = <math>2\cos(4x)</math> and solving gets 194.
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 +
== 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>
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<b>Remarks about <math>\boldsymbol{(*)}</math></b>
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<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 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)
 +
 
 +
== 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

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 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

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 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

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 (Solution)

~MRENTHUSIASM (Minor Formatting)

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 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

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 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

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 (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
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All AMC 10 Problems and Solutions

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