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

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Notice that <math>(a^{k} + a^{-k})^2 = a^{2k} + a^{-2k} + 2</math>. Thus <math>a^4 + a^{-4} = (a^2 + a^{-2})^2 - 2 = [(a + a^{-1})^2 - 2]^2 - 2 = 194\ \mathrm{(D)}</math>.
 
Notice that <math>(a^{k} + a^{-k})^2 = a^{2k} + a^{-2k} + 2</math>. Thus <math>a^4 + a^{-4} = (a^2 + a^{-2})^2 - 2 = [(a + a^{-1})^2 - 2]^2 - 2 = 194\ \mathrm{(D)}</math>.
  
=== Solution 2 ===
+
=== Solution 2(LIFEHACK) ===
<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>.  
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Notice that <math>(a^{4} + a^{-4}) = (a^{2} + a^{-2} + 2</math>. Since \ \mathrm{(D)}<math> 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 ===
 +
</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>.  
  
 
=== Solution 3 ===
 
=== Solution 3 ===
 
(similar to Solution 1)
 
(similar to Solution 1)
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 ===
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>.
+
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>
+
</math>S_4=\boxed{194}<math>
  
 
=== Solution 5 ===
 
=== Solution 5 ===
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.
+
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)$ and solving gets 194.
  
 
== See also ==
 
== See also ==

Revision as of 22:12, 4 January 2020

Problem

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

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


Solutions

Solution 1

Notice that $(a^{k} + a^{-k})^2 = a^{2k} + a^{-2k} + 2$. Thus $a^4 + a^{-4} = (a^2 + a^{-2})^2 - 2 = [(a + a^{-1})^2 - 2]^2 - 2 = 194\ \mathrm{(D)}$.

Solution 2(LIFEHACK)

Notice that $(a^{4} + a^{-4}) = (a^{2} + a^{-2} + 2$. Since \ \mathrm{(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 ===$4a = a^2 + 1$. We apply the [[quadratic formula]] to get$a = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.

Thus$ (Error compiling LaTeX. ! Missing $ inserted.)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 3 === (similar to Solution 1) We know that$ (Error compiling LaTeX. ! Missing $ inserted.)a+\frac{1}{a}=4$. We can square both sides to get$a^2+\frac{1}{a^2}+2=16$, so$a^2+\frac{1}{a^2}=14$. Squaring both sides again gives$a^4+\frac{1}{a^4}+2=14^2=196$, so$a^4+\frac{1}{a^4}=\boxed{194}$.

=== Solution 4 === We let$ (Error compiling LaTeX. ! Missing $ inserted.)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 5 === 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 (ProblemsAnswer KeyResources)
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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