The number of ordered triples (x,y,z) satisfy 3x^2+3y^2+z^2-2xy+2yz=0 is?

1 Answer
Jul 28, 2018

ONE

Explanation:

3x^2+3y^2+z^2-2xy+2yz=0

=>2x^2+y^2+(x^2-2xy+y^2)+(y^2+2yz+z^2)=0

=>2x^2+y^2+(x-y)^2+(y+z)^2=0

Hence for real values of x,y,z to satisfy the above relation we have

x=0,y=0,x=yand y=-z

Hence x=y=z=0

So only ordered triplet is (0,0,0)

So the number of ordered triplet is ONE