Difference between revisions of "2021 AMC 10B Problems/Problem 15"
(→Solution 2) |
(→Solution 2) |
||
Line 19: | Line 19: | ||
&=(-2x^7+3x^6)+x^3 \ | &=(-2x^7+3x^6)+x^3 \ | ||
&=(x^6-2x^5)+x^3 \ | &=(x^6-2x^5)+x^3 \ | ||
− | &=(-x^5+x^4+x^3) | + | &=(-x^5+x^4+x^3) \ |
− | &=-x^3(x^2-x-1) | + | &=-x^3(x^2-x-1) = \boxed{(\textbf{B}) 0} |
− | |||
\end{align*}</cmath> | \end{align*}</cmath> | ||
~Lcz | ~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