Difference between revisions of "2007 AMC 10A Problems/Problem 20"
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=== Solution 3 === | === Solution 3 === | ||
| − | We know that <math>a+\frac{1}{a}=4</math>. We can square both sides to get <math>a^2+\frac{1}{a^2}+2=16</math>, so <math>a^2+\frac{1}{a^2}=14</math>. Squaring both sides again gives <math>a^4+\frac{1}{a^4}+2=14^2=196</math>, so <math>a^4+\frac{1}{a^4}=\boxed{194}</math> | + | We know that <math>a+\frac{1}{a}=4</math>. We can square both sides to get <math>a^2+\frac{1}{a^2}+2=16</math>, so <math>a^2+\frac{1}{a^2}=14</math>. Squaring both sides again gives <math>a^4+\frac{1}{a^4}+2=14^2=196</math>, so <math>a^4+\frac{1}{a^4}=\boxed{194}</math>. |
=== Solution 4 === | === Solution 4 === | ||
Revision as of 16:59, 31 January 2016
Problem
Suppose that the number 
satisfies the equation 
. What is the value of 
?
Solution
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
. We apply the quadratic formula to get 
.
Thus 
(so it doesn't matter which root of we use). Using the binomial theorem we can expand this out and collect terms to get 
.
Solution 3
We know that 
. We can square both sides to get 
, so 
. Squaring both sides again gives 
, so 
.
Solution 4
We let and 
be roots of a certain quadratic. Specifically 
. We use Newton's Sums given the coefficients to find 
.
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. 
