2020 AMC 10A Problems/Problem 14

Revision as of 13:49, 1 February 2020 by Emathmaster (talk | contribs) (Solution 2)

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

$\textbf{- Emathmaster}$

Video Solution

https://youtu.be/ZGwAasE32Y4

~IceMatrix

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 10 Problems and Solutions

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