Question 372828
Extraneous solutions occur because we perform extraneous operations.    For example , solve {{{2+5/(x-4) = (x+1)/(x-4)}}}.  Some would mulitply both sides by x-4 (which is an extraneous operation).  This would give 2(x-4)+5 = x+1,
2x - 8 + 5 = x+1, which would give x = 4.  BUT we know that x = 4 is not in the domain of the equation, hence cannot be a solution. 
Another equation is {{{2x = 1 - sqrt(2-x)}}}.  What we normally do is square both sides after transposing 1 to the other side (which is essentially an extraneous operation).  It would give the equation {{{(2x-1)^2 = 2-x}}}, which after simplification gives{{{4x^2-3x-1 = 0}}}, and the candidates -1/4 and 1.  Unfortunately, 1 does not satisfy the original equation (-1/4 does!)  The possible root x = 1 was "introduced" in the part where we squared both sides of the equation.  
Hope you understood that, student...