Question 173483
{{{sqrt(x+10)+2=x}}}


I think you probably had the process correct, but you made a sign error.


Add -2 to both sides:


{{{sqrt(x+10)=x-2}}}


Square both sides:


{{{x+10=x^2-4x+4}}}


Collect terms and put in standard form:


{{{x^2-5x-6=0}}}


{{{(1)(-6)=-6}}} and {{{(1)+(-6)=-5}}} so:


{{{(x-6)(x+1)=0}}}


{{{x-6=0}}} => {{{x=6}}} or


{{{x+1=0}}} => {{{x=-1}}}


{{{6}}} is a valid root because {{{sqrt(6+10)+2=sqrt(16)+2=4+2=6}}} but


{{{-1}}} is NOT a valid root because {{{sqrt(-1+10)+2=sqrt(9)+2=3+2=5<>-1}}}


The invalid, or extraneous root was introduced by the act of squaring the variable in the process of solving the equation.  You must ALWAYS check your answers against the original equation any time you square a variable in the process of solving.