I've seen problems like this before on RU-vid. The symmetry of the problem suggests that x and y are conjugates, i.e. x=a+b and y=a-b. Trying that: x+y = a+b+a-b = 2a = 8 So: a = 4 Then: xy = a^2 - b^2 = 16 - b^2 = 48 Rearranging and simplifying: b^2 = -32 So: b = ±4i√2 Then: x = a+b = 4+4i√2 or 4-4i√2 And: y = 8-x = 4-4i√2 or 4+4i√2 I enjoy your interesting problems.
With Vieta, we know, that those x and y in (x+y=-b=8 and x*y=c=48) are just the solutions of z^2 + bz + c = 0, so you immediately get 4:14 z^2-8z+48=0 (x or z doesn't matter, I didn't want to reuse x!) => z_12= 4 +/- 4 sqrt(2) i. Thus x_1 = z_1 and y_1 = z_2 || x_2 = z_2 and y_2 = z_1, that is x is one of the two z solutions and y is the other one.