2007 AMC 10A Problems/Problem 20
Problem
Suppose that the number 
satisfies the equation 
. What is the value of 
?
Solutions
Solution 1
Notice that 
. Thus ![$a^4 + a^{-4} = (a^2 + a^{-2})^2 - 2 = [(a + a^{-1})^2 - 2]^2 - 2 = 194\ \mathrm{(D)}$](http://latex.artofproblemsolving.com/f/b/4/fb4f4375486400cfb7604312fe3981dd7f3d3f59.png)
.
Solution 2(LIFEHACK)
Notice that 
. Since \ \mathrm{(D)}
4a = a^2 + 1![$. We apply the [[quadratic formula]] to get$](http://latex.artofproblemsolving.com/9/a/4/9a4d9da471029a97d6d025492586b6fb31f33e0f.png)
a = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.
Thus$ (Error compiling LaTeX. Unknown error_msg)a^4 + a^{-4} = (2+\sqrt{3})^4 + \frac{1}{(2+\sqrt{3})^4} = (2+\sqrt{3})^4 + (2-\sqrt{3})^4
a![$we use). Using the [[binomial theorem]] we can expand this out and collect terms to get$](http://latex.artofproblemsolving.com/2/a/a/2aa97a30ba39df6812b2e5db5fb907938a84754c.png)
194$.
=== Solution 3 ===
(similar to Solution 1)
We know that$ (Error compiling LaTeX. Unknown error_msg)a+\frac{1}{a}=4
a^2+\frac{1}{a^2}+2=16
a^2+\frac{1}{a^2}=14
a^4+\frac{1}{a^4}+2=14^2=196a^4+\frac{1}{a^4}=\boxed{194}$.
=== Solution 4 ===
We let$ (Error compiling LaTeX. Unknown error_msg)a
1/a
x^2-4x+1=0![$. We use [[Newton's Sums]] given the coefficients to find$](http://latex.artofproblemsolving.com/c/6/9/c6988f47d0b0f368cc597c91c63b034330349700.png)
S_4
S_4=\boxed{194}
a
\cos(x)
i\sin(x)
a + a^{-1} = 2\cos(x)
\cos(x) = 2![$. Then by [[De Moivre's Theorem]],$](http://latex.artofproblemsolving.com/f/7/f/f7f995e5ff83743aa0f2fb8d474b6294d2853576.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
