2021 AMC 10B Problems/Problem 15
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
Solution 5
The equation we are given is Yuck. Fractions and radicals! We multiply both sides by
square, and re-arrange to get
Now, let us consider the expression we wish to acquire. Factoring out
we have
Then, we notice that
Furthermore,
Thus, our answer is
Video Solution by OmegaLearn (Algebraic Manipulations and Symmetric Polynomials)
~ pi_is_3.14
Video Solution by Interstigation (Simple Silly Bashing)
~ Interstigation
Video Solution by TheBeautyofMath
Not the most efficient method, but gets the job done.
https://youtu.be/L1iW94Ue3eI?t=1468
~IceMatrix
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.