Question 122042
I presume you solved this in the following fashion:


{{{sqrt(x+2)-x=0}}}
{{{sqrt(x+2)=x}}}
Square both sides:
{{{x+2=x^2}}}
{{{x^2-x-2=0}}}
{{{(x-2)(x+1)=0}}}


{{{x=2}}} or {{{x=-1}}}


The reason that {{{x=-1}}} doesn't check ({{{sqrt(-1+2)-(-1)=2<>0}}}) is that you introduced an extraneous root when you squared both sides of the equation in the process of solving it.  {{{2}}} is the only element of the solution set of the original equation.


The following is a trivial example of what happens when you square both sides of an equation:


Let {{{x=1}}}


Square both sides:


{{{x^2=1^2=1}}}


{{{x^2-1=0}}}


{{{(x-1)(x+1)=0}}}


{{{x=1}}} or {{{x=-1}}}


{{{x=1}}} is fine, but {{{x=-1}}}, by substitution in the original equation, leads us to the absurdity that {{{-1=1}}}.  Therefore {{{x=-1}}} is an extraneous root introduced in the process of squaring both sides of the equation and it must be excluded from the solution set.