Difference between revisions of "2007 AMC 10A Problems/Problem 20"
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=== Solution 2(LIFEHACK) === | === Solution 2(LIFEHACK) === | ||
− | Notice that <math>(a^{4} + a^{-4}) = (a^{2} + a^{-2} + 2</math>. Since | + | Notice that <math>(a^{4} + a^{-4}) = (a^{2} + a^{-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. | PS: Because this is a multiple choice test, this works. | ||
=== Solution 3 === | === Solution 3 === | ||
− | < | + | <math>4a = a^2 + 1</math>. We apply the [[quadratic formula]] to get <math>a = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}</math>. |
− | Thus < | + | Thus <math>a^4 + a^{-4} = (2+\sqrt{3})^4 + \frac{1}{(2+\sqrt{3})^4} = (2+\sqrt{3})^4 + (2-\sqrt{3})^4</math> (so it doesn't matter which root of <math>a</math> we use). Using the [[binomial theorem]] we can expand this out and collect terms to get <math>194</math>. |
=== Solution 3 === | === Solution 3 === | ||
(similar to Solution 1) | (similar to Solution 1) | ||
− | We know that < | + | 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 === | ||
− | We let < | + | We let <math>a</math> and <math>1/a</math> be roots of a certain quadratic. Specifically <math>x^2-4x+1=0</math>. We use [[Newton's Sums]] given the coefficients to find <math>S_4</math>. |
− | < | + | <math>S_4=\boxed{194}</math> |
=== Solution 5 === | === Solution 5 === | ||
− | Let < | + | 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. |
== See also == | == See also == |
Revision as of 22:13, 4 January 2020
Problem
Suppose that the number satisfies the equation . What is the value of ?
Solutions
Solution 1
Notice that . Thus .
Solution 2(LIFEHACK)
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 3
(similar to Solution 1) 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 .
Solution 5
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 |
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