Question

Answer with steps plz​

Answer 1

Hint

• Polynomial identity

$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(xy+yz+zx)$

Solution

Given, $\dfrac{x^{2}+y^{2}+z^{2}-100}{xy+yz+zx} =-2$

$\implies x^2+y^2+z^2-100=-2(xy+yz+zx)$

$\implies x^2+y^2+z^2+2(xy+yz+zx)=100$

Given, $x+y=3z$

$\implies x+y+z=4z$

$\implies x^{2}+y^{2}+z^{2}+2(xy+yz+zx)=16z^2$

Finally, $16z^2=100$

$\implies z^2=\dfrac{100}{16}$

$\implies z=\pm \dfrac{10}{4} =\boxed{\pm\dfrac{5}{2} }$