Difference between revisions of "2020 AMC 10A Problems/Problem 14"
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− | ==Solution 4== | + | ==Solution 4 (Bashing)== |
This is basically bashing using Vieta's formulas to find <math> x </math> and <math> y </math> (which I highly do not recommend, I only wrote this solution for fun). | This is basically bashing using Vieta's formulas to find <math> x </math> and <math> y </math> (which I highly do not recommend, I only wrote this solution for fun). | ||
We use Vieta's to find a quadratic relating <math> x </math> and <math> y </math>. We set <math> x </math> and <math> y </math> to be the roots of the quadratic <math> Q ( n ) = n^2 - 4n - 2 </math> (because <math> x + y = 4 </math>, and <math> xy = -2 </math>). We can solve the quadratic to get the roots <math> 2 + \sqrt{6} </math> and <math> 2 - \sqrt{6} </math>. <math> x </math> and <math> y </math> are "interchangeable", meaning that it doesn't matter which solution <math> x </math> or <math> y </math> is, because it'll return the same result when plugged in. So we plug in <math> 2 + \sqrt{6} </math> for <math> x </math> and <math> 2 - \sqrt{6} </math> and get <math> \boxed{\textbf{(D)}\ 440} </math> as our answer. | We use Vieta's to find a quadratic relating <math> x </math> and <math> y </math>. We set <math> x </math> and <math> y </math> to be the roots of the quadratic <math> Q ( n ) = n^2 - 4n - 2 </math> (because <math> x + y = 4 </math>, and <math> xy = -2 </math>). We can solve the quadratic to get the roots <math> 2 + \sqrt{6} </math> and <math> 2 - \sqrt{6} </math>. <math> x </math> and <math> y </math> are "interchangeable", meaning that it doesn't matter which solution <math> x </math> or <math> y </math> is, because it'll return the same result when plugged in. So we plug in <math> 2 + \sqrt{6} </math> for <math> x </math> and <math> 2 - \sqrt{6} </math> and get <math> \boxed{\textbf{(D)}\ 440} </math> as our answer. | ||
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~Baolan | ~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 <math>4 + \frac{x^5 + y^5}{x^2 y^2}</math>, because <math>x + y = 4</math>. This is equal to <math>4 + \frac{(x + y)(x^4 - x^3 y + x^2 y^2 - x y^3 + y^4)}{4} = x^4 + y^4 - x^3 y - x y^3 + 8</math>. We can factor and reduce <math>x^4 + y^4</math> to <math>(x^2 + y^2)^2 - 2 x^2 y^2 = ((x + y)^2 - 2xy)^2 - 8 = 400 - 8 = 392</math>. Now our expression is just <math>400 - (x^3 y + x y^3)</math>. We factor <math>x^3 y + x y^3</math> to get <math>(xy)(x^2 + y^2) = -40</math>. So the answer would be <math>400 - (-40) | ||
+ | = \boxed{\textbf{(D)} 440} </math>. | ||
+ | |||
+ | |||
+ | ~ Baolan | ||
==Video Solution== | ==Video Solution== |
Revision as of 11:50, 3 February 2020
Real numbers and
satisfy
and
. What is the value of
- Note: All solutions involve factoring
Contents
[hide]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
.
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
.
~ Baolan
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.