<s>[INST] 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.
 [/INST] 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 $x=-2$    
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