2020 AMC 10A Problems/Problem 14

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 $\frac{(20)(88)}{4}=\boxed{\textbf{(D)}\ 440}$ .

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

Solution 7

As before, simplify the expression to $x + y + \frac{x^5 + y^5}{x^2y^2}.$ Since $x + y = 4$ and $x^2y^2 = 4$ , we substitute that in to obtain $4 + \frac{x^5 + y^5}{4}.$ Now, we must solve for $x^5 + y^5$ . Start by squaring $x + y$ , to obtain $x^2 + 2xy + y^2 = 16$ Simplifying, $x^2 + y^2 = 20$ . Squaring once more, we obtain $x^4 + y^4 + 2x^2y^2 = 400$ Once again simplifying, $x^4 + y^4 = 392$ . Now, to obtain the fifth powers of $x$ and $y$ , we multiply both sides by $x + y$ . We now have $x^5 + x^4y + xy^4 + y^5 = 1568$ , or $x^5 + y^5 + xy(x^3 + y^3) = 1568$ We now solve for $x^3 + y^3$ . $(x + y)^3=x^3 + y^3 + 3xy(x + y) = 64$ , so $x^3 + y^3 = 88$ . Plugging this back into $x^5 + x^4y + xy^4 + y^5 = 1568$ , we find that $x^5 + y^5 = 1744$ , so we have $4 + \frac{1744}{4}.$ . This equals 440, so our answer is $\boxed{\textbf{(D)} 440}$ .

~Binderclips1

Solution 8

We can use Newton Sums to solve this problem. We start by noticing that we can rewrite the equation as $\frac{x^3}{y^2} + \frac{y^3}{x^2} + x + y.$ Then, we know that $x + y = 4,$ so we have $\frac{x^3}{y^2} + \frac{y^3}{x^2} + 4.$ We can use the equation $x \cdot y = -2$ to write $x = \frac{-2}{y}$ and $y = \frac{-2}{x}.$ Next, we can plug in these values of $x$ and $y$ to get $\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5}{4} + \frac{y^5}{4},$ which is the same as $\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5 + y^5}{4}.$ Then, we use Newton sums where $S_n$ is the elementary symmetric sum of the sequence and $P_n$ is the Newton sum ( $x^n + y^n$ ). Using this, we can make the following Newton sums: $P_1 = S_1$ $P_2 = P_1 S_1 - 2S_2$ $P_3 = P_2 S_1 - P_1 S_2$ $P_4 = P_3 S_1 - P_2 S_2$ $P_5 = P_4 S_1 - P_3 S_2.$ We also know that $S_1$ is 4 because $x + y$ is four, and we know that $S_2$ is $-2$ because $x \cdot y$ is $-2$ as well. Then, we can plug in values! We have $P_1 = S_1 = 4$ $P_2 = P_1 S_1 - 2S_2 = 16 - (-4) = 20$ $P_3 = P_2 S_1 - P_1 S_2 = 80 - (-8) = 88$ $P_4 = P_3 S_1 - P_2 S_2 = 88 \cdot 4 - (-40) = 392$ $P_5 = P_4 S_1 - P_3 S_2 = 392 \cdot 4 - (-2) \cdot 88 = 1744.$ We earlier noted that $\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^5 + y^5}{4},$ so we have that this equals $\frac{1744}{4},$ or $436.$ Then, plugging this back into the original equation, this is $436 + 4$ or $440,$ so our answer is $\boxed{\textbf{(D)}\ 440}.$

~Coolpeep

Video Solution

https://youtu.be/ZGwAasE32Y4

~IceMatrix

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