Difference between revisions of "2020 AMC 10A Problems/Problem 14"
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Revision as of 12:41, 3 August 2020
Contents
Problem
Real numbers and
satisfy
and
. What is the value of
Solution
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
.
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
Video Solution
~IceMatrix
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.