Difference between revisions of "2021 AMC 10B Problems/Problem 15"
(→Solution 2) |
(→Solution 2) |
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&=(-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 22:16, 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