Hello from MrBillDoesMath!
Answer:
(x,y) = (2 - 2 i sqrt(2), 2 + 2sqrt(2))
(x,y) = (2 + 2 i sqrt(2), 2 - 2sqrt(2))
Discussion:
x + y = 4 => y = 4 - x
Substituting the y value in xy = 12 gives
xy = x(4-x) = 12 =>
4x - x^2 = 12 =>
x^2 -4x + 12 = 0 => (applying the quadratic formula)
x = 2 - 2 i sqrt(2) ( y = 4 - x = 4 - (2 -2isqrt(2) )
and
x = 2 + 2 i sqrt(2) ( y = 4 - x = 4 - (2 +2isqrt(2) )
Thank you,
MrB
