Difference between revisions of "2007 AMC 10A Problems/Problem 20"
MRENTHUSIASM (talk | contribs) |
MRENTHUSIASM (talk | contribs) (→Solution 1 (Generalized)) |
||
Line 9: | Line 9: | ||
a^4 + a^{-4} &= \left(a^2 + a^{-2}\right)^2 - 2 \\ | a^4 + a^{-4} &= \left(a^2 + a^{-2}\right)^2 - 2 \\ | ||
&= \left[\left(a + a^{-1}\right)^2 - 2\right]^2 - 2 \\ | &= \left[\left(a + a^{-1}\right)^2 - 2\right]^2 - 2 \\ | ||
− | &= \boxed{\text{(D)}\ 194}.</cmath> | + | &= \boxed{\text{(D)}\ 194}. |
+ | \end{align*}</cmath> | ||
~Azjps (Fundamental Logic) | ~Azjps (Fundamental Logic) | ||
Revision as of 23:32, 24 June 2021
Contents
Problem
Suppose that the number satisfies the equation
. What is the value of
?
Solution 1 (Generalized)
Notice that for all real numbers we have
from which
We apply this result twice to get the answer:
~Azjps (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2
Notice that . 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 3
. 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 4
(similar to Solution 1)
We know that . We can square both sides to get
, so
. Squaring both sides again gives
, so
.
Solution 5
We let and
be roots of a certain quadratic. Specifically
. We use Newton's Sums given the coefficients to find
.
Solution 6
Let =
+
. Then
so
. Then by De Moivre's Theorem,
=
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.