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

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

Find

Please explain every part thoroughly. Thank you!

Guest Jan 31, 2015

#2
+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:

#1
+80874
+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
+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
#3
+80874
0

Very nice, Bertie....

CPhill  Feb 1, 2015

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