Roots in equation In the equation
$\sqrt{x-2} - \sqrt{6x-11} + \sqrt{x+3} =0$
I got the roots of $x$ being $6$ and $7\sqrt{3}$.
Considering the graph shows only $6$ as being a valid solution, how should I go as figuring this out in the equation itself?

Square both sides
$$\sqrt{x-2} + \sqrt{x+3} = \sqrt{6x-11} $$
$$2x+1 + 2\sqrt{(x-2)(x+3)}=6x-11$$
$$\sqrt{(x-2)(x+3)}=2x-6$$
Square again
$$x^2+x-6 = 4x^2 - 24x+36$$
$$ 3x^2 - 25x+42=0$$
$$ (x-6)(3x-7)=0$$
$$x=6,\>\>\>x=\frac 73$$
Plug into the original equation and check validity. Only $x=6$ is the true solution. (Edit: See comment below by @Joe as to how the spurious solution $x=\frac 73$ came about.)
