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Difference between revisions of "2007 AMC 10A Problems/Problem 20"

m (→‎Solution 1 (Generalized))
(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)
  
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== Solution 2 ==
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== 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>
 +
 
 +
~Rbhale12 (Fundamental Logic)
 +
 
 +
~MRENTHUSIASM (Reconstruction)
 +
 
 +
== 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.
  
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== Solution 3 ==
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== 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 4 ==
 
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(similar to Solution 1)
 
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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 5 ==
 
== Solution 5 ==

Revision as of 23:00, 24 June 2021

Problem

Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$. What is the value of $a^{4} + a^{ - 4}$?

$\text{(A)}\ 164 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 192 \qquad \text{(D)}\ 194 \qquad \text{(E)}\ 212$

Solution 1 (Direct)

Notice that for all real numbers $k,$ we have $a^{2k} + a^{-2k} + 2 = \left(a^{k} + a^{-k}\right)^2,$ from which \[a^{2k} + a^{-2k} = \left(a^{k} + a^{-k}\right)^2-2.\] We apply this result twice to get the answer: \begin{align*} a^4 + a^{-4} &= \left(a^2 + a^{-2}\right)^2 - 2 \\ &= \left[\left(a + a^{-1}\right)^2 - 2\right]^2 - 2 \\ &= \boxed{\text{(D)}\ 194}. \end{align*} ~Azjps (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 2 (Stepwise)

Squaring both sides of $a+a^{-1}=4$ gives $a^2+a^{-2}+2=16,$ from which $a^2+a^{-2}=14.$

Squaring both sides of $a^2+a^{-2}=14$ gives $a^4+a^{-4}+2=196,$ from which $a^4+a^{-4}=\boxed{\text{(D)}\ 194}.$

~Rbhale12 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 3

Notice that $(a^{4} + a^{-4}) = (a^{2} + a^{-2})^{2} - 2$. 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

$4a = a^2 + 1$. We apply the quadratic formula to get $a = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.

Thus $a^4 + a^{-4} = (2+\sqrt{3})^4 + \frac{1}{(2+\sqrt{3})^4} = (2+\sqrt{3})^4 + (2-\sqrt{3})^4$ (so it doesn't matter which root of $a$ we use). Using the binomial theorem we can expand this out and collect terms to get $194$.

Solution 5

We let $a$ and $1/a$ be roots of a certain quadratic. Specifically $x^2-4x+1=0$. We use Newton's Sums given the coefficients to find $S_4$. $S_4=\boxed{194}$

Solution 6

Let $a$ = $\cos(x)$ + $i\sin(x)$. Then $a + a^{-1} = 2\cos(x)$ so $\cos(x) = 2$. Then by De Moivre's Theorem, $a^4 + a^{-4}$ = $2\cos(4x)$ 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

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