Find the real solutions of the equations:$x+y=2$, $xy-z^2=1$ Can anyone help with this? I will show you what I did.

Find the real solutions of the equation:$$x+y=2\tag{1}$$ $$xy-z^2=1 \tag{2}$$

From equation (2), I have $$xy=1+z^2 \tag{3}$$ From equations (1) and (3), I have a new equation $$r^2-2r+(1+z^2)=0$$ The roots of this equation are the solutions. How do I find the root?

Given $x+y=2$ and $xy-z^2=1$
So $xy=1+z^2\geq 1>0$.
So  $x,y>0$
Using $\bf{A.M\geq G.M},$ we get $$\frac{x+y}{2}\geq \sqrt{xy}\Rightarrow (x+y)^2\geq 4xy$$ and equality hold when $x=y$
So $4\geq 4xy\Rightarrow xy\leq 1\Rightarrow 1+z^2\leq 1\Rightarrow z^2\leq 0\Rightarrow z=0$
and $x=y=1.$ So we get $(x,y,z) = (1,1,0)$
