2007 AMC 10A Problems/Problem 20
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 \ \mathrm{(D)}4a = a^2 + 1a = \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})^4a194$.
=== Solution 3 === (similar to Solution 1) We know that$ (Error compiling LaTeX. ! Missing $ inserted.)a+\frac{1}{a}=4a^2+\frac{1}{a^2}+2=16a^2+\frac{1}{a^2}=14a^4+\frac{1}{a^4}+2=14^2=196a^4+\frac{1}{a^4}=\boxed{194}$.
=== Solution 4 === We let$ (Error compiling LaTeX. ! Missing $ inserted.)a1/ax^2-4x+1=0S_4S_4=\boxed{194}a\cos(x)i\sin(x)a + a^{-1} = 2\cos(x)\cos(x) = 2a^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 | |
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All AMC 10 Problems and Solutions |
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