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Find all ordered pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 64 + xy$.

 May 27, 2024
 #1
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x + y = 10         square both sides

 

x^2 + 2xy + y^2 = 100

 

x^2 + y^2  = 100 - 2xy

 

So

 

100 - 2xy  = 64 + xy

 

100 - 64 = 3xy

 

36 = 3xy

 

12 = xy →   y = 12/x

 

So

 

x + 12/x = 10

 

x^2 -10x + 12  =  0

 

x = [ 10 +/- sqrt [100 - 48] ] / [ 2  =  [ 10 +/- sqrt [ 52] ]/ 2  =  [10 +/- 2sqrt13 ] / 2  = 5 +/- sqrt 13

 

Using the conjugate property

 

(x,y) =  ( 5 + sqrt 13 , 5 -sqrt 13 )  ,  (5 -sqrt 13 , 5 + sqrt 13)

 

cool cool cool

 May 28, 2024

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