Question

Find solutions of equation x^2+y^2+z^2=8x-2z-8 if x is natural number and y zEZ.

Answer 1

Step-by-step explanation:

We have,

${x}^{2} + {y}^{2} + {z}^{2} = 8x - 2z - 8$

$\implies {x}^{2} - 8x + {y}^{2} + {z}^{2} + 2z + 8 = 0$

$\implies {x}^{2} - 8x + 16 + {y}^{2} + {z}^{2} + 2z + 1 - 1 + 8 - 16 = 0$

$\implies (x - 4)^{2} + {y}^{2} + (z + 1)^{2} + 7 - 16 = 0$

$\implies (x - 4)^{2} + {y}^{2} + (z + 1)^{2} = 9$

We are given, $x\in\mathbb{N}$ and $y,z\in\mathbb{Z}$

So,

One solution will be $x=1,y=0,z=-1$

Answer 2

Given:

$x=4-\sqrt{8-y^{2}-z^{2}-2 z} \text { or } x=4+\sqrt{8-y^{2}-z^{2}-2 z}$

$\text { Solve the equations: } x=4-\sqrt{8-y^{2}-z^{2}-2 z} \text { or } x=4+\sqrt{8-y^{2}-z^{2}-2 z}$

$\text { Answer: } x=4-\sqrt{8-y^{2}-z^{2}-2 z} \text { or } x=4+\sqrt{8-y^{2}-z^{2}-2 z}$