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Find all ordered pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 32 + 2xy$.
For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(2,4),(-3,9)" (without the quotation marks).

 Jun 8, 2024
 #1
avatar+129399 
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x + y  = 10

y = 10 -x

 

x^2 + (10 -x)^2  = 32 + 2x(10-x)

 

x^2 + x^2 -20x + 100 = 32 + 20x -2x^2

 

4x^2 -40x + 68 = 0

 

x^2 -10x + 17  = 0

 

x^2 -10x  =  -17

 

x^2 -10x + 25  = -17 + 25

 

(x -5)^2  = 8

 

x -5 = +/- sqrt (8) = +/- 2sqrt 2

 

x = 5 +  2sqrt2      or      x =  5 - 2sqrt 2

 

y = 10 -x   =    5 -2sqrt 2    or   5 + 2sqrt 2

 

 

(x,y)  = (5 + 2sqrt 2 , 5-2sqrt 2)   or ( 5- 2sqrt 2, 5 + 2sqrt 2)

 

 

cool cool cool

 Jun 8, 2024

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