Difference between revisions of "2021 AMC 10B Problems/Problem 15"
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Multiplying both sides by <math>x</math> and using the quadratic formula, we get <math>\frac{\sqrt{5} \pm 1}{2}</math>. We can assume that it is <math>\frac{\sqrt{5}+1}{2}</math>, but notice that this is also a solution the equation <math>x^2-x-1=0</math>, i.e. we have <math>x^2=x+1</math>. Repeatedly using this on the given (you can also just note Fibonacci numbers), | Multiplying both sides by <math>x</math> and using the quadratic formula, we get <math>\frac{\sqrt{5} \pm 1}{2}</math>. We can assume that it is <math>\frac{\sqrt{5}+1}{2}</math>, but notice that this is also a solution the equation <math>x^2-x-1=0</math>, i.e. we have <math>x^2=x+1</math>. Repeatedly using this on the given (you can also just note Fibonacci numbers), | ||
<cmath> \begin{align*} | <cmath> \begin{align*} | ||
− | (x^11)-7x^7+x^3 &= (x^10+x^9)-7x^7+x^3 \ | + | (x^{11})-7x^7+x^3 &= (x^{10}+x^9)-7x^7+x^3 \ |
&=(2x^9+x^8)-7x^7+x^3 \ | &=(2x^9+x^8)-7x^7+x^3 \ | ||
&=(3x^8+2x^7)-7x^7+x^3 \ | &=(3x^8+2x^7)-7x^7+x^3 \ | ||
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&=\boxed{(\textbf{B}) 0} | &=\boxed{(\textbf{B}) 0} | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
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+ | ~Lcz |
Revision as of 21:16, 11 February 2021
Problem
The real number satisfies the equation . What is the value of
Solution 1
We square to get . We subtract 2 on both sides for and square again, and see that so . We can divide our original expression of by to get that it is equal to . Therefore because is 7, it is equal to .
Solution 2
Multiplying both sides by and using the quadratic formula, we get . We can assume that it is , but notice that this is also a solution the equation , i.e. we have . Repeatedly using this on the given (you can also just note Fibonacci numbers),
~Lcz