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Find all pairs (x,y) of real numbers such that x + y = 10 and x^2 + y^2 = 56 + 2xy.
For example, to enter the solutions (2,4) and (-3,9), you would enter "(2,4),(-3,9)" (without the quotation marks).

 Apr 30, 2022
 #1
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x + y =  10

Square both sides

x^2 + 2xy + y^2 =  100

x^2 + y^2  = 100 - 2xy        and          x^2 + y^2  =  56 + 2xy

 

Setting the last wo equal we have that

 

100  - 2xy =  56 + 2xy

 

100  -  56 =  4xy

 

44 =  4xy

 

11 =  xy

 

y =  11 / x

 

x +  11/x  =  10

 

x^2 + 11  =  10x

 

x^2 - 10x + 11   =  0

 

x^2  - 10x + 25  =  -11  + 25

 

(x - 5)^2  =  14

 

x - 5  =   5 +sqrt (14)      or   5 - sqrt (14)

 

By the conjugate property of quadratic roots

When  x = the first answer  y=  5 -sqrt 14

When x = the second answer y = 5 + sqrt 14

 

So

 

(x,y)   = ( 5 + sqrt 14 , 5 -sqrt 14)   or  (5 -sqrt 14, 5 + sqrt 14)

 

 

cool cool cool

 Apr 30, 2022

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