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

Revision as of 10:24, 20 February 2020

Problem

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

$x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}$

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 (Bashing)

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

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 $4 + \frac{x^5 + y^5}{x^2 y^2}$ , because $x + y = 4$ . This is equal to $4 + \frac{(x + y)(x^4 - x^3 y + x^2 y^2 - x y^3 + y^4)}{4} = x^4 + y^4 - x^3 y - x y^3 + 8$ . We can factor and reduce $x^4 + y^4$ to $(x^2 + y^2)^2 - 2 x^2 y^2 = ((x + y)^2 - 2xy)^2 - 8 = 400 - 8 = 392$ . Now our expression is just $400 - (x^3 y + x y^3)$ . We factor $x^3 y + x y^3$ to get $(xy)(x^2 + y^2) = -40$ . So the answer would be $400 - (-40) = \boxed{\textbf{(D)} 440}$ .

Solution 6 (Complete Binomial Theorem)

We first simplify the expression to $x + y + \frac{x^5 + y^5}{x^2y^2}.$ Then, we can solve for $x$ and $y$ given the system of equations in the problem. Since $xy = -2,$ we can substitute $\frac{-2}{x}$ for $y$ . Thus, this becomes the equation $x - \frac{2}{x} = 4.$ Multiplying both sides by $x$ , we obtain $x^2 - 2 = 4x,$ or $x^2 - 4x - 2 = 0.$ By the quadratic formula we obtain $x = 2 \pm \sqrt{6}$ . We also easily find that given $x = 2 \pm \sqrt{6}$ , $y$ equals the conjugate of $x$ . Thus, plugging our values in for $x$ and $y$ , our expression equals $4 + \frac{(2 - \sqrt{6})^5 + (2 + \sqrt{6})^5}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}$ By the binomial theorem, we observe that every second terms of the expansions $x^5$ and $y^5$ 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 $x^5 + y^5$ . Thus, our expression equals $4 + \frac{2(2^5 + \tbinom{5}{2}2^3 \times 6 + \tbinom{5}{4}2 \times 36)}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}.$ which equals $4 + \frac{2(872)}{4}$ which equals $\boxed{\textbf{(D)} 440}$ .

~ fidgetboss_4000

Video Solution

https://youtu.be/ZGwAasE32Y4

~IceMatrix

@@ Line 48: / Line 48: @@
 Multiplying both sides by <math>x</math>, we obtain <math>x^2 - 2 = 4x,</math> or
 <cmath>x^2 - 4x - 2 = 0.</cmath>
-By the quadratic formula we obtain <math>x = 2 \pm 6</math>.
+By the quadratic formula we obtain <math>x = 2 \pm \sqrt{6}</math>.
-We also easily find that given <math>x = 2 \pm 6</math>, <math>y</math> equals the conjugate of <math>x</math>.
+We also easily find that given <math>x = 2 \pm \sqrt{6}</math>, <math>y</math> equals the conjugate of <math>x</math>.
 Thus, plugging our values in for <math>x</math> and <math>y</math>, our expression equals
 <cmath>4 + \frac{(2 - \sqrt{6})^5 + (2 + \sqrt{6})^5}{(2 - \sqrt{6})^2(2 + \sqrt{6})^2}</cmath>

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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