Question

If (x2 +y2 + z2 -100)/ (xy + yz+ zx )= -2 and x+y = 3z then find the value of z.

Answer 1

Hi,

$$\begin{gathered} \frac{{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2-100}{\mathrm{xy}+\mathrm{yz}+\mathrm{zx}}=-2 \\ {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2-100=-2\mathrm{xy}-2\mathrm{yz}-2\mathrm{zx} \\ {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2+2\mathrm{xy}+2\mathrm{yz}+2\mathrm{zx}=100 \\ {\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)}^2=100 \\ \mathrm{x}+\mathrm{y}+\mathrm{z}=\sqrt{100}=10\mathrm{or}-10 \\ \mathrm{also}\mathrm{x}+\mathrm{y}=3\mathrm{z} \\ \mathrm{so}3\mathrm{z}+\mathrm{z}=10\mathrm{or}-10 \\ 4\mathrm{z}=10\mathrm{or}-10 \\ \mathrm{z}=2.5\mathrm{or}-2.5 \end{gathered}$$

Regards