Equations problem

**Issoka** · July 15th 2010, 07:24 PM

HI! Actually I have no idea of how to solve this problem, could any one help me please?

You have to find de values for A, B and C.

Thanks

**Prove It** · July 15th 2010, 08:06 PM

You have

$\begin{cases}x + y + z = 3 \\ x^2 + y^2 + z^2 = 9 \\ xyz = -2\end{cases}$

From equation 1:

$x = 3 - y - z$ , so

$\begin{cases}(3 - y - z)^2 + y^2 + z^2 = 9\\ (3 - y - z)yz = -2\end{cases}$

$\begin{cases}(3 - y - z)^2 + y^2 + z^2 = 9 \\ zy^2 + (z^2 - 3z)y - 2 = 0\end{cases}$

Since the final equation is now a quadratic in $y$ , you can solve for $y$ in terms of $z$ using the quadratic formula.

Everything else should piece together after that.

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