Solve the system of equations: x + y + z = a x2 + y2 + z2 = b2 xy = z2 where a and b are constants. Give the conditions that a and b must satisfy so that x; y; z (the solutions of the system) are distinct positive numbers.

x+y+z=a  ---(1)
x^2+y^2+z^2=b^2 ---(2)
xy=z^2 ---(3)
squaring on both sides on first equation
(x+y+z)^2=a^2
x^2+y^2+z^2+2(xy+yz+zx)=a^2
b^2+2(xy+yz+zx)=a^2
2(xy+yz+zx)=a^2 - b^2
2(z^2+yz+zx)=a^2-b^2
2z(z+x+y)=a^2-b^2
2az=a^2-b^2
z=(a^2-b^2)/2a
x^2+y^2= -(a^2-b^2)^2/4a^2
(x-y)^2=x^2+y^2-2xy
            = -3(a^2-b^2)^2/4a^2
x-y=sqrt( -3(a^2-b^2)^2/4a^2)=sqrt(-3) a^2-b^2/2a  ----(5)
x+y=a-z
x+y=a^2+b^2/2a ---(6)
 by adding 5 and 6 we get x
x=(a^2+b^2)/4a + sqrt(3) * i * (a^2-b^2)/4a  where i is imaginary
by subtractiing 5 from 6
y=(a^2+b^2)/4a - sqrt(3) * i * (a^2-b^2)/4a

the values of x,y,z are
x=(a^2+b^2)/4a + sqrt(3) * i * (a^2-b^2)/4a
y=(a^2+b^2)/4a - sqrt(3) * i * (a^2-b^2)/4a

z=(a^2-b^2)/2a

i got to this solution i hope this helps