Difference between revisions of "2021 AMC 10B Problems/Problem 15"
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==Solution 3== | ==Solution 3== | ||
− | We can immediately note that the exponents of <math>x^{11}-7x^7+x^3</math> are an arithmetic sequence, so they are symmetric around the middle term. So, <math>x^{11}-7x^7+x^3 = x^7(x^4-7+\frac{1}{x^4})</math>. We can see that since <math>x+\frac{1}{x} = \sqrt{5}</math>, <math>x^2+2+\frac{1}{x^2} = 5</math> and therefore <math>x^2+\frac{1}{x^2} = 3</math>. Continuing from here, we get <math>x^4+2+\frac{1}{x^4} = 9</math>, so <math>x^4-7+\frac{1}{x^4} = 0</math>. We don't even need to find what <math>x^3</math> is! This is since <math>x^3\cdot0</math> is evidently <math>\boxed{ | + | We can immediately note that the exponents of <math>x^{11}-7x^7+x^3</math> are an arithmetic sequence, so they are symmetric around the middle term. So, <math>x^{11}-7x^7+x^3 = x^7(x^4-7+\frac{1}{x^4})</math>. We can see that since <math>x+\frac{1}{x} = \sqrt{5}</math>, <math>x^2+2+\frac{1}{x^2} = 5</math> and therefore <math>x^2+\frac{1}{x^2} = 3</math>. Continuing from here, we get <math>x^4+2+\frac{1}{x^4} = 9</math>, so <math>x^4-7+\frac{1}{x^4} = 0</math>. We don't even need to find what <math>x^3</math> is! This is since <math>x^3\cdot0</math> is evidently <math>\boxed{(B) 0}</math>, which is our answer. |
~sosiaops | ~sosiaops |
Revision as of 19:56, 18 February 2021
Contents
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
, and 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
Solution 3
We can immediately note that the exponents of are an arithmetic sequence, so they are symmetric around the middle term. So,
. We can see that since
,
and therefore
. Continuing from here, we get
, so
. We don't even need to find what
is! This is since
is evidently
, which is our answer.
~sosiaops
Solution 4
We begin by multiplying by
, resulting in
. Now we see this equation:
. The terms all have
in common, so we can factor that out, and what we're looking for becomes
. Looking back to our original equation, we have
, which is equal to
. Using this, we can evaluate
to be
, and we see that there is another
, so we put substitute it in again, resulting in
. Using the same way, we find that
is
. We put this into
, resulting in
, so the answer is
.
~purplepenguin2
Video Solution by OmegaLearn (Algebraic Manipulations and Symmetric Polynomials)
~ pi_is_3.14
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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All AMC 10 Problems and Solutions |