Solve $\begin{cases} x + y + z = 2 \\ 2xy - z^2 = 4 \\ \end{cases} $ for x, y, z.

Question

It came to my mind to rewrite the expression above as $$\begin{cases} x + y = 2 - z \\ 2xy = (2 - z)^2 + 4z \\ \end{cases} $$ and see if there any restrictions on the values of the variables occur.

What I can see is that $(2 - z)^2 + 4z \gt 0$ for all $z$ (since $z^2 > -4$) but when I express $y$ in terms of $z$ and $x$ $$\begin{cases} y = 2 - z - x\\ x(x - (2 - z)) = (2 - z)^2 + 4z \\ \end{cases} $$ I see that $x(x - (2 - z)) \gt 0$ not for all $z$. This is all I got for now.

The answer is $(2, 2, -2)$.

Answer 1

You have: $$ x + y= 2-z\\ xy = 2 + \frac{z^2}{2}. $$ From Vieta's formulas, $x$ and $y$ are roots of $$ t^2 - (2-z)t + 2 + \frac{z^2}{2} = 0; $$ hence $$ t = \frac{2 - z \pm \sqrt{(2-z)^2 - 8 - 2z^2}}{2} = \frac{2 - z \pm \sqrt{-(z+2)^2}}{2}. $$ So, we have $z=-2$, and $t=2$; therefore $t=x=y=2$ and $z=-2$.

Greate thanks to @Shailesh for notes.

Answer 2

$\bf{My\; Solution::}$ Given $$x+y+z = 2$$ and $$2xy-z^2 = 4$$

Now Put $z=2-(x+y)$ into $2xy-z^2=4\Rightarrow 2xy-[2-(x+y)]^2=4$

So we get $$2xy-[4+(x+y)^2-4(x+y)]=4$$

So we get $$2xy-4-x^2-y^2-2xy+4x+4y=4$$

So we get $$x^2+y^2-4x-4y+8=0$$

So we get $$(x-2)^2+(y-2)^2 = 0\;,$$ Which is possible for real no. when

$(x-2)=0$ and $(y-2) = 0$ , So we get $x=2$ and $y=2$ and $z=-2$