Difference between revisions of "2020 AMC 10A Problems/Problem 14"
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~IceMatrix | ~IceMatrix | ||
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== See Also == | == See Also == | ||
{{AMC10 box|year=2020|ab=A|num-b=13|num-a=15}} | {{AMC10 box|year=2020|ab=A|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:30, 15 December 2020
Contents
Problem
Real numbers and
satisfy
and
. What is the value of
Solutions
Solution 1
Continuing to combine
From the givens, it can be concluded that
. Also,
This means that
. Substituting this information into
, we have
. ~PCChess
Solution 2
As above, we need to calculate . Note that
are the roots of
and so
and
. Thus
where
and
as in the previous solution. Thus the answer is
. Note(
, and
)
Solution 3
Note that Now, we only need to find the values of
and
Recall that and that
We are able to solve the second equation, and doing so gets us
Plugging this into the first equation, we get
In order to find the value of we find a common denominator so that we can add them together. This gets us
Recalling that
and solving this equation, we get
Plugging this into the first equation, we get
Solving the original equation, we get ~emerald_block
Solution 4 (Bashing)
This is basically bashing using Vieta's formulas to find and
(which I highly do not recommend, I only wrote this solution for fun).
We use Vieta's to find a quadratic relating and
. We set
and
to be the roots of the quadratic
(because
, and
). We can solve the quadratic to get the roots
and
.
and
are "interchangeable", meaning that it doesn't matter which solution
or
is, because it'll return the same result when plugged in. So we plug in
for
and
and get
as our answer.
~Baolan
Solution 5 (Bashing Part 2)
This usually wouldn't work for most problems like this, but we're lucky that we can quickly expand and factor this expression in this question.
We first change the original expression to , because
. This is equal to
. We can factor and reduce
to
. Now our expression is just
. We factor
to get
. So the answer would be
.
Solution 6 (Complete Binomial Theorem)
We first simplify the expression to
Then, we can solve for
and
given the system of equations in the problem.
Since
we can substitute
for
.
Thus, this becomes the equation
Multiplying both sides by
, we obtain
or
By the quadratic formula we obtain
.
We also easily find that given
,
equals the conjugate of
.
Thus, plugging our values in for
and
, our expression equals
By the binomial theorem, we observe that every second terms of the expansions
and
will cancel out (since a positive plus a negative of the same absolute value equals zero). We also observe that the other terms not canceling out are doubled when summing the expansions of
.
Thus, our expression equals
which equals
which equals
.
~ fidgetboss_4000
Solution 7
As before, simplify the expression to
Since
and
, we substitute that in to obtain
Now, we must solve for
. Start by squaring
, to obtain
Simplifying,
. Squaring once more, we obtain
Once again simplifying,
. Now, to obtain the fifth powers of
and
, we multiply both sides by
.
We now have
, or
We now solve for
.
, so
.
Plugging this back into
, we find that
, so we have
. This equals 440, so our answer is
.
~Binderclips1
Solution 8
We can use Newton Sums to solve this problem.
We start by noticing that we can rewrite the equation as
Then, we know that
so we have
We can use the equation
to write
and
Next, we can plug in these values of
and
to get
which is the same as
Then, we use Newton sums where
is the elementary symmetric sum of the sequence and
is the power sum (
). Using this, we can make the following Newton sums:
We also know that
is 4 because
is four, and we know that
is
because
is
as well.
Then, we can plug in values! We have
We earlier noted that
so we have that this equals
or
Then, plugging this back into the original equation, this is
or
so our answer is
~Coolpeep
Video Solution
Education, The Study of Everything
~IceMatrix
~savannahsolver
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.