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?

Let $x-2=p^2,x+3=q^2;p,q\ge0$
As $x+3>x-2, \dfrac pq<1\ \ (1)$
$6x-11=a(x-2)+b(x+3)$
$x=2\implies1=5b$
$x+3=0\implies6(-3)-11=a(-3-2)$
$$\implies p+q=\sqrt{\dfrac{q^2+29p^2}5}$$
$$25(p+q)^2=q^2+29p^2$$
$$2p^2-25pq+12q^2=0$$
$$\dfrac pq=\dfrac{25\pm\sqrt{25^2-4\cdot2\cdot12}}{2\cdot2}=\dfrac{25\pm23}4$$
$\implies \dfrac pq =\dfrac12$ by $(1)$
Now take square in both sides and replace the values of $p,q$
