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

m (→‎Solution 4 (Solves for a))
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<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
a^4+a^{-4}&=\left(2+\sqrt{3}\right)^4 + \frac{1}{\left(2+\sqrt{3}\right)^4} \\
 
a^4+a^{-4}&=\left(2+\sqrt{3}\right)^4 + \frac{1}{\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{\text{(D)}\ 194}.
 
&=\boxed{\text{(D)}\ 194}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>

Revision as of 15:28, 25 June 2021

Problem

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

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

Solution 1 (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 $a^4+a^{-4}=\boxed{\text{(D)}\ 194}.$

~Rbhale12 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 2 (Decreases the Powers)

Notice that for all real numbers $k,$ we have $a^{2k} + a^{-2k} + 2 = \left(a^{k} + a^{-k}\right)^2,$ from which \[a^{2k} + a^{-2k} = \left(a^{k} + a^{-k}\right)^2-2.\] We apply this result twice to get the answer: \begin{align*} a^4 + a^{-4} &= \left(a^2 + a^{-2}\right)^2 - 2 \\ &= \left[\left(a + a^{-1}\right)^2 - 2\right]^2 - 2 \\ &= \boxed{\text{(D)}\ 194}. \end{align*} ~Azjps (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 3 (Binomial Theorem)

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

We raise both sides of $a+a^{-1}=4$ to the fourth power, then apply the Binomial Theorem: \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{\text{(D)}\ 194}. \end{align*} ~MRENTHUSIASM

Solution 4 (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 $a=2\pm\sqrt3.$

Note that the roots are reciprocals of each other. Therefore, choosing either value for $a$ gives the same value for $a^4+a^{-4}:$ \begin{align*} a^4+a^{-4}&=\left(2+\sqrt{3}\right)^4 + \frac{1}{\left(2+\sqrt{3}\right)^4} \\ &=\left(2+\sqrt{3}\right)^4+\left(2-\sqrt{3}\right)^4 &&&(*) \\ &=\boxed{\text{(D)}\ 194}. \end{align*} Remarks in $\boldsymbol{(*)}$

  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 5 (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{\text{(D)}\ 194}. \end{alignat*} ~Albert1993 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 6 (Answer Choices)

Notice that \[a^{4} + a^{-4} = \left(a^{2} + a^{-2}\right)^{2} - 2.\] 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 $\boxed{\text{(D)}\ 194}.$

~Thanosaops (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

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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