Mathematics - Quadratic Equation Question with Solution | TestHub

$\because x+y+z=1$ and $2xy-z^2=1$

$\because AM\ge GM$

$\Rightarrow \frac{x+y}{2}\ge \sqrt{(xy)}\Rightarrow \left(\frac{1-z}{2}\right)\ge \sqrt{\left(\frac{1+z^2}{2}\right)}\Rightarrow (1-z{)}^2\ge 2\left(1+z^2\right)$ $\Rightarrow z^2+2z+1\le 0\Rightarrow (z+1{)}^2\le 0$

$\therefore z+1=0\Rightarrow z=-1$

Then $x+y=2$ and $xy=1$

Hence $x=y=1$