Difference between revisions of "2007 AMC 10A Problems/Problem 20"
MRENTHUSIASM (talk | contribs) m (→Solution 1 (Generalized)) |
MRENTHUSIASM (talk | contribs) (In the next few edits, I will reformat the solutions and rearrange them by educational values. If anyone disagrees with me, please PM me--I am sure that we can figure out a solution.) |
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~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
− | == Solution 2 == | + | == Solution 2 (Stepwise) == |
+ | Squaring both sides of <math>a+a^{-1}=4</math> gives <math>a^2+a^{-2}+2=16,</math> from which <math>a^2+a^{-2}=14.</math> | ||
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+ | Squaring both sides of <math>a^2+a^{-2}=14</math> gives <math>a^4+a^{-4}+2=196,</math> from which <math>a^4+a^{-4}=\boxed{\text{(D)}\ 194}.</math> | ||
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+ | ~Rbhale12 (Fundamental Logic) | ||
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+ | ~MRENTHUSIASM (Reconstruction) | ||
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+ | == Solution 3 == | ||
Notice that <math>(a^{4} + a^{-4}) = (a^{2} + a^{-2})^{2} - 2</math>. Since D is the only option 2 less than a perfect square, that is correct. | Notice that <math>(a^{4} + a^{-4}) = (a^{2} + a^{-2})^{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 | + | == Solution 4 == |
<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>. | <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 <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>. | 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>. | ||
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== Solution 5 == | == Solution 5 == |
Revision as of 23:00, 24 June 2021
Contents
[hide]Problem
Suppose that the number satisfies the equation . What is the value of ?
Solution 1 (Direct)
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 (Stepwise)
Squaring both sides of gives from which
Squaring both sides of gives from which
~Rbhale12 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 3
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 4
. 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 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 |
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