Roots in equation

Question

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?

Answer 1

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.)

Answer 2

Let $x=7\sqrt 3$ into $$\sqrt{x-2} - \sqrt{6x-11} + \sqrt{x+3} $$ and you get $$ \sqrt{x-2} - \sqrt{6x-11} + \sqrt{x+3} \approx -0.7869 \ne 0$$

Thus $x=7\sqrt 3$ is not a solution.

Answer 3

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 $p,q$