Difference between revisions of "2020 AMC 10A Problems/Problem 14"
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== Solution 2 == | == Solution 2 == | ||
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>. | ||
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| + | \textbf{- Emathmaster} | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 13:48, 1 February 2020
Real numbers 
and 
satisfy 
and 
. What is the value of![\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?\]](http://latex.artofproblemsolving.com/1/b/7/1b77f92edbab9a6134e14217ee0c56700cef3dbf.png)
Contents
Solution
![\[x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}\]](http://latex.artofproblemsolving.com/3/c/b/3cb720eb6b236c7897b32cbe34768bfa695b2ba3.png)
Continuing to combine ![\[\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2y^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}\]](http://latex.artofproblemsolving.com/7/4/4/74411ff57b4c958d41ca7cb6c6d0230b331b0ad7.png)
From the givens, it can be concluded that 
. Also, ![\[(x+y)^2=x^2+2xy+y^2=16\]](http://latex.artofproblemsolving.com/f/e/4/fe4d79118d31b37765c0ae83dc534680c73f2ee4.png)
This means that 
. Substituting this information into 
, we have 
. ~PCChess
Solution 2
As above, we need to calculate 
. Note that 
are the roots of 
and so 
and 
. Thus 
where and as in the previous solution. Thus the answer is 
.
\textbf{- Emathmaster}
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
