Question 175240
It really depends on whether you mean {{{sqrt(x)+2-x=0}}} or {{{sqrt(x + 2) - x=0}}}.  You could also have meant {{{sqrt(x + 2 -x)=0}}} but that reduces to the absurdity that {{{sqrt(2) =0}}}, so I am discounting that possibility.


Case 1:


{{{sqrt(x)+2-x=0}}}


Let {{{t=sqrt(x)}}}, then {{{t + 2 - t^2=0}}}


In standard form:  {{{t^2 - t - 2 = 0}}}.


This factors to:  {{{(t - 2)(t + 1)=0}}} so {{{t = 2}}} or {{{t = -1}}}


But {{{t=sqrt(x)}}} so {{{sqrt(x) = 2}}} or {{{sqrt(x) = -1}}} meaning:


{{{x = 4}}} or {{{x = 1}}}


However, since we converted to a higher degree equation in order to solve this, we may have introduced an extraneous (and therefore incorrect) root.  Check:


{{{sqrt(4) +  2 - 4 = 2 + 2 - 4 = 0}}}  Checks.


{{{sqrt(1) +  2 - 1 = 2 <> 0}}}  Does not check.  Extraneous root.


Therefore the solution set is {{{x = 4}}}


Case 2:


{{{sqrt(x + 2) - x=0}}}


Add {{{x}}} to both sides:


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


Square both sides:


{{{x + 2 = x^2}}}


Standard Form:


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


Factor:


{{{(x - 2)(x + 1)= 0}}}


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


Again, since we squared both sides in the solution process, we have the possibility of an extraneous root.


Check:

{{{sqrt(2 + 2) - 2= 2 - 2 = 0}}} Checks


{{{sqrt(-1 + 2) - (-1) = 1 + 1 = 2 <> 0}}} Does not check.  Extraneous root.


Solution set is {{{x = 2}}}