Originally Posted by

**SheekhKebab** An interesting solution:-

x+y+z=4 and x^{2}+y^{2}+z^{2}=6,

y+z=4-x ...............(i)

and y^{2}+z^{2}=6-x^{2}

yz=$\frac{1}{2}$[(y+z)^{2}-(y^{2}+z^{2})]

yz=x^{2}-4x+5,.................(ii)

Hence, y and z are roots of:

t^{2}-(4-x)t + (x^{2}-4x+5)=0

Since the roots y and z are real, hence D>=0:

i.e (4-x)^{2}-4(x^{2}-4x+5)>=0

or (3x-2)(x-2)<=0

x = [ $\frac{2}{3}$,2]

and by symmetry y and z are also [$\frac{2}{3}$,2]

Now, you can clearly see that the Discriminant indeed gives the range of solution set x,y,z. Or am I missing something ?