+0  
 
0
277
3
avatar

Let x and y be real numbers whose absolute values are different and that satisfy

\begin{align*} x^3 &= 20x + 7y <br /> y^3 &= 7x + 20y \end{align*}

Find xy.

 

 

Please explain every part thoroughly. Thank you!

difficulty advanced
Guest Jan 31, 2015

Best Answer 

 #2
avatar+889 
+10

Let xy=k, so that y = k/x.

Substitute this into the first equation to get  $$x^{4}-20x^{2}-7k=0.$$

Solve this (using the usual formula) as a quadratic in x squared to get $$x^{2}=10\pm\sqrt{100+7k}$$.

The original equations are symmetric in x and y so repeating the procedure will produce an identical result for y.

Since x and y are different, one will have the positive sign in the middle, the other the negative sign.

Multiplying the two results produces (difference between two squares)

$$x^{2}y^{2}=100-(100+7k)=-7xy.$$

Since $$xy\ne 0,$$ it follows that xy = -7.

Bertie  Feb 1, 2015
Sort: 

3+0 Answers

 #1
avatar+81029 
+10

x^3 = 20x + 7y   (1)

y^3 = 7x + 20y   (2)  subtract (2) from(1)

x^3 - y^3 = 13(x-y)    factor the left side

(x-y)(x^2 + xy + y^2) = 13(x-y)     divide through by (x-y)

x^2 + xy + y^2) = 13   subtract 13 from both sides

x^2 + xy +y^2 - 13 =  0     (3) ....  this is a rotated ellipse....

We need to find the intersection points of (1),(2) and (3)

Here's a graph

Graph

The intersection points we're after are where the absolute values of x and y are different

These occur at (-4.14, 1.691), (4.14, -1.691),(1.691, -4.14) and (-1.691, 4.14)

And xy will  be -(1.691)(4.14)  which will be ≈ -7

BTW... here are the exact values generated by WolframAlpha

http://www.wolframalpha.com/input/?i=solve+x^3+%3D+20x+%2B+7y%2C+y^3+%3D+7x+%2B+20y%2Cx^2+%2B+xy+%2By^2+-+13+%3D++0+

I have a feeling that the math was pretty sticky getting these.....!!! 

 

CPhill  Jan 31, 2015
 #2
avatar+889 
+10
Best Answer

Let xy=k, so that y = k/x.

Substitute this into the first equation to get  $$x^{4}-20x^{2}-7k=0.$$

Solve this (using the usual formula) as a quadratic in x squared to get $$x^{2}=10\pm\sqrt{100+7k}$$.

The original equations are symmetric in x and y so repeating the procedure will produce an identical result for y.

Since x and y are different, one will have the positive sign in the middle, the other the negative sign.

Multiplying the two results produces (difference between two squares)

$$x^{2}y^{2}=100-(100+7k)=-7xy.$$

Since $$xy\ne 0,$$ it follows that xy = -7.

Bertie  Feb 1, 2015
 #3
avatar+81029 
0

Very nice, Bertie....

CPhill  Feb 1, 2015

9 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details