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

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

Revision as of 13:49, 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

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

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 (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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