Is -2 is a root of the equation : $\sqrt {x^2 - 8} = \sqrt {3x + 2}$? Is -2 is a root of the equation : $\sqrt{x^2 - 8} = \sqrt{3x + 2}$ ?
Is it any limitation of the root of this equation?
If the root is -2, both sides of the equation are equal to 2i.
Is this acceptable since the variable x has no limitation.
 A: The equation on its own is not really well-posed; you need to say which kind of number you allow $x$ to be, and what you mean by the symbol $\sqrt{\phantom{X}}$.
If the question is "Find all real $x$ which satisfy the equation, where $\sqrt{\phantom{X}}$ denotes the usual real square root function", then the answer is no, since $\sqrt{-2}$ is undefined.
But if the question is "Find all complex $x$ which satisfy the equation, where $\sqrt{\phantom{X}}$ denotes the principal branch of the complex square root function" (or some other branch, but this must be specified whenever one talks about complex square roots), then the answer is yes.
In some cases, it might be clear from context which alternative is meant. For example, if you're given this question in a real-variable calculus course which doesn't explain branch cuts for complex square roots, then it's probably the first answer ("no") that's expected.
A: No, the square root is not defined for negative arguments.
A: Notice, the given equation is $\sqrt{x^2-8}=\sqrt{3x+2}$, now squaring both the sides we get $$x^2-8=3x+2$$ $$ x^2-3x-10=0$$ $$(x+2)(x-5)=0$$ On solving we get $$x=-2, \ 5$$
Now, let's check out both the roots, if they are acceptable by satisfying the given (original) equation  
Substituting $x=-2$, we get $LHS=RHS=\sqrt{-2}$ which is undefined value hence, the $x=-2$ is unacceptable i.e. it's not a root of the given equation. 
While, Substituting $x=5$, we get $LHS=RHS=\sqrt{17}$ which is defined value hence, the $x=5$ is a root of given equation it is not $x=-2$    
A: In addition to just substituing and checking, we can show this analytically.
$$\sqrt{x^2-8}=\sqrt{3x+2} \implies x^2-8=3x+2$$
$$\implies x^2-3x-10=0$$
By quadratic formula:
$$\implies x=-2 \text{ or } x=5$$
It is worth noting that in the form you originally wrote the equation, there would be a problem of arguments in the square root. By allowing the squaring that we do here, we are agreeing to overlook this, since the square of a purely imaginary number is real.
A: We find solution roots unwittingly( unintentionally as extraneous / spurious)  included for $ \sqrt{x^2 - 8} = \pm \sqrt{3x + 2} $ due to squarings.
So if roots in complex domain/range are  considered, then $ x= - 2 $ is allowable in complex range. I.e., if solution of $ \sqrt{z^2 - 8} = \pm \sqrt{3 z + 2} $ is asked, then it is acceptable as a root. The x=5 real solution root anyhow.
