262i could reach only till the point that
the soln will lie on -> (x+1)2+(y+1)2=1
$\textbf{Solve for real system of equation}\\\\ $\mathbf{3.(x^2+y^2+z^2)-2(xy+yz+zx)=1}$\\\\ $\mathbf{x+y-z=-1}$
-
UP 0 DOWN 0 0 4
4 Answers
262
21then obviously there r infinite solutions
as we can see only 2 eqns r there
so 99% possibility we hav to get some sum of square = zero
but we r not getting that...
11\sum_{cyc}(x^2-2xy+y^2)=1-(x^2+y^2+z^2)
Now from the second eqn, we get
1-(x^2+y^2+z^2)=2xy-2xz-2yz
From which we get
2(x-y)^2+2z^2=0
Thus the soln
x=y=\frac{-1}{2}
& z=0
