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

(In the next few edits, I will reformat the solutions and rearrange them by educational values. If anyone disagrees with me, please PM me--I am sure that we can figure out a solution.)
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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>
 
<math>\text{(A)}\ 164 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 192 \qquad \text{(D)}\ 194 \qquad \text{(E)}\ 212</math>
  
−
== Solution 1 (Direct) ==
+
== Solution 1 (Decreases the Powers) ==
 
Notice 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:
 
Notice 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:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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~MRENTHUSIASM (Reconstruction)
 
~MRENTHUSIASM (Reconstruction)
  
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== Solution 2 (Stepwise) ==
+
== 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>
  
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~MRENTHUSIASM (Reconstruction)
 
~MRENTHUSIASM (Reconstruction)
 +
 +
== Solution 3 (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>
 +
 +
We raise both sides of <math>a+a^{-1}=4</math> to the fourth power, then apply the Binomial Theorem:
 +
<cmath>\begin{align*}
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\binom40a^4a^0+\binom41a^3a^{-1}+\binom42a^2a^{-2}+\binom43a^1a^{-3}+\binom44a^0a^{-4}&=256 \\
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a^4+4a^2+6+4a^{-2}+a^{-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(14)&=250 \\
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a^4+a^{-4}&=\boxed{\text{(D)}\ 194}.
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\end{align*}</cmath>
 +
~MRENTHUSIASM
  
 
== Solution 3 ==
 
== Solution 3 ==

Revision as of 23:42, 24 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 (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 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 $a^4+a^{-4}=\boxed{\text{(D)}\ 194}.$

~Rbhale12 (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 3

Notice that $(a^{4} + a^{-4}) = (a^{2} + a^{-2})^{2} - 2$. Since D is the only option 2 less than a perfect square, that is correct.

PS: Because this is a multiple choice test, this works.

Solution 4

$4a = a^2 + 1$. We apply the quadratic formula to get $a = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.

Thus $a^4 + a^{-4} = (2+\sqrt{3})^4 + \frac{1}{(2+\sqrt{3})^4} = (2+\sqrt{3})^4 + (2-\sqrt{3})^4$ (so it doesn't matter which root of $a$ we use). Using the binomial theorem we can expand this out and collect terms to get $194$.

Solution 5

We let $a$ and $1/a$ be roots of a certain quadratic. Specifically $x^2-4x+1=0$. We use Newton's Sums given the coefficients to find $S_4$. $S_4=\boxed{194}$

Solution 6

Let $a$ = $\cos(x)$ + $i\sin(x)$. Then $a + a^{-1} = 2\cos(x)$ so $\cos(x) = 2$. Then by De Moivre's Theorem, $a^4 + a^{-4}$ = $2\cos(4x)$ and solving gets 194.

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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