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2007 AMC 10A Problems/Problem 20

Revision as of 20:04, 13 July 2020 by Edward4zheng (talk | contribs) (→‎Solution 2)

Problem

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

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


Solutions

Solution 1

Notice that $(a^{k} + a^{-k})^2 = a^{2k} + a^{-2k} + 2$. Thus $a^4 + a^{-4} = (a^2 + a^{-2})^2 - 2 = [(a + a^{-1})^2 - 2]^2 - 2 = 194\ \mathrm{(D)}$.

Solution 2

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. LOL

Solution 3

$4a = a^2 + 1$. 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 3

(similar to Solution 1) We know that $a+\frac{1}{a}=4$. We can square both sides to get $a^2+\frac{1}{a^2}+2=16$, so $a^2+\frac{1}{a^2}=14$. Squaring both sides again gives $a^4+\frac{1}{a^4}+2=14^2=196$, so $a^4+\frac{1}{a^4}=\boxed{194}$.

Solution 4

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 5

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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