Question

x+y+z=0, xy+yz+zx=1/3, x^2+y^2+z^2=-1/3, x:y:z=?​

Answer 1

Given Data is Wrong  , x+y+z=0, xy+yz+zx=1/3, x^2+y^2+z^2=-1/3, x:y:z=?​

Step-by-step explanation:

(x + y + z) = 0

Squaring both sides

=> (x + y + z)² = 0²

=> x² + y² + z² + 2(xy + yz + zx ) = 0

=> -1/3 + 2(1/3)  =0

=> 1/3  = 0

Which is not True

=> Data Given is wrong

Similar Question if

(x + y + z) = 1   & xy+yz+zx=1/3,   then find x:y:z

=> x² + y² + z² + 2(xy + yz + zx ) = 1

=> x² + y² + z²  =  1/3

    xy+yz+zx = 1/3

Which will hold true if

x = y = z

=> x : y : z ::: 1: 1: 1