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Difference between revisions of "2020 AMC 10A Problems/Problem 14"

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    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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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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Revision as of 23:42, 1 February 2020

Real numbers $x$ and $y$ satisfy $x + y = 4$ and $x \cdot y = -2$. What is the value of\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?\] $\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480$

Solution 1

\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}\]

Continuing to combine \[\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}\] From the givens, it can be concluded that $x^2y^2=4$. Also, \[(x+y)^2=x^2+2xy+y^2=16\] This means that $x^2+y^2=20$. Substituting this information into $\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}$, we have $\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{\textbf{(D)}\ 440}$. ~PCChess

Solution 2

As above, we need to calculate $\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}$. Note that $x,y,$ are the roots of $x^2-4x-2$ and so $x^3=4x^2+2x$ and $y^3=4y^2+2y$. Thus $x^3+y^3=4(x^2+y^2)+2(x+y)=4(20)+2(4)=88$ where $x^2+y^2=20$ and $x^2y^2=4$ as in the previous solution. Thus the answer is $\frac{(20)(88)}{4}=\boxed{\textbf{(D)}\ 440}$.

$\textbf{- Emathmaster}$

Solution 3

Note that $( x^3 + y^3 ) ( \frac{1}{y^2} + \frac{1}{x^2} ) = x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y.$ Now, we only need to find the values of $x^3 + y^3$ and $\frac{1}{y^2} + \frac{1}{x^2}.$

Recall that $x^3 + y^3 = (x + y) (x^2 - xy + y^2),$ and that $x^2 - xy + y^2 = (x + y)^2 - 3xy.$ We are able to solve the second equation, and doing so gets us $4^2 - 3(-2) = 22.$ Plugging this into the first equation, we get $x^3 + y^3 = 4(22) = 88.$

In order to find the value of $\frac{1}{y^2} + \frac{1}{x^2},$ we find a common denominator so that we can add them together. This gets us $\frac{x^2}{x^2 y^2} + \frac{y^2}{x^2 y^2} = \frac{x^2 + y^2}{(xy)^2}.$ Recalling that $x^2 + y^2 = (x+y)^2 - 2xy$ and solving this equation, we get $4^2 - 2(-2) = 20.$ Plugging this into the first equation, we get $\frac{1}{y^2} + \frac{1}{x^2} = \frac{20}{(-2)^2} = 5.$

Solving the original equation, we get $x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y = (88)(5) = \boxed{\textbf{(D)}\ 440}.$ ~emerald_block


Solution 4

This is basically bashing using Vieta's formulas to find $x$ and $y$ (which I highly do not recommend, I only wrote this solution for fun).


We use Vieta's to find a quadratic relating $x$ and $y$. We set $x$ and $y$ to be the roots of the quadratic $Q ( n ) = n^2 - 4n - 2$ (because $x + y = 4$, and $xy = -2$). We can solve the quadratic to get the roots $2 + \sqrt{6}$ and $2 - \sqrt{6}$. $x$ and $y$ are "interchangeable", meaning that it doesn't matter which solution $x$ or $y$ is, because it'll return the same result when plugged in. So we plug in $2 + \sqrt{6}$ for $x$ and $2 - \sqrt{6}$ and get $\boxed{\textbf{(D)}\ 440}$ as our answer.


~Baolan

Video Solution

https://youtu.be/ZGwAasE32Y4

~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

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