Question

If $\frac{x^2+y^2+z^2-64}{xy-yz-zx}=-2$ and $x+y=3z$, then the value of $z$ is

• A. $2$
• B. $3$
• C. $4$
• D. $-2$

Answer 1

The correct option is C $4$

$\frac{x^2{+y}^2{+z}^2-64}{xy-yz-zx}=-2$
$\Rightarrow x^2+y^2+z^2=-2xy+2yz+2zx+64$
Now, ${(x+y+z)}^2=x^2{+y}^2{+z}^2+2xy+2yz+2zx$
$\Rightarrow {(3z+z)}^2=-2xy+2yz+2zx+64+2xy+2yz+2zx$
$\Rightarrow 16z^2=4yz+4xz+64$

$\Rightarrow 16z^2=4z(x+y)+64$
$\Rightarrow 16z^2=4z\left(3z\right)+64$

$\Rightarrow 16z^2-12z^2=64$
$\Rightarrow 4z^2=64$

$\Rightarrow z^2=16$

$\Rightarrow z=4$