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

Revision as of 23:51, 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 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.

Solution 7 (Answer Choices)

Notice that $a^{4} + a^{-4} = \left(a^{2} + a^{-2}\right)^{2} - 2,$ from which the answer must be $2$ less than a perfect square. The only possibility is $\boxed{\text{(D)}\ 194}.$

@@ Line 36: / Line 36: @@
 \end{align*}</cmath>
 ~MRENTHUSIASM
-== Solution 3 ==
-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.
-PS: Because this is a multiple choice test, this works.
 == Solution 4 ==
@@ Line 53: / Line 48: @@
 == Solution 6 ==
 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.
+== Solution 7 (Answer Choices) ==
+Notice that <math>a^{4} + a^{-4} = \left(a^{2} + a^{-2}\right)^{2} - 2,</math> from which the answer must be <math>2</math> less than a perfect square. The only possibility is <math>\boxed{\text{(D)}\ 194}.</math>
 == 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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