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).
+130081 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)
