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

(→‎Solution 7 (Answer Choices))
(Removed the complex number solution. Originally it said cos x = 2, which is always false. You are welcome to put it back if you can come up with the right logic. Sorry.)
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<math>S_4=\boxed{194}</math>
 
<math>S_4=\boxed{194}</math>
  
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== Solution 6 ==
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== Solution 6 (Answer Choices) ==
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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 7 (Answer Choices) ==
 
 
Notice 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{\text{(D)}\ 194}.</math>
 
Notice 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{\text{(D)}\ 194}.</math>
  

Revision as of 00:55, 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 (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 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 (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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