Question 79742
The best way to understand what extraneous roots are is to go through a problem where you get one. The following is a solution to a problem that I did previously which dealt with extraneous roots:

{{{sqrt(2y+7)+4=y}}}
First, get the radical expression alone on one side of the equation:
{{{sqrt(2y+7)=y-4}}}  
Now, square both sides of the equation to get rid of the radical:
{{{2y+7=(y-4)^2}}}
Simplify the right hand side:
{{{2y+7=y^2-8y+16}}}
Combine like terms and equate to zero:
{{{y^2-10y+9=0}}}
Factor the expression:
{{{(y-9)(y-1)=0}}}
Equate each factor to zero, and solve:
{{{y=9}}} and {{{y=1}}}

Now, this part is important...you need to plug these answers back in to
the original equation to be sure they are not "extraneous." An extraneous
root may be mathematically correct, but it is not the true answer. If you
plug y=1 back into the original equation, you will see that the equation
DOES NOT hold true. Hence, this is an extraneous root. If you plug y=9 back
into the original equation, you will see that the equation DOES hold true,
so that one is your answer.

Good Luck,
tutor_paul@yahoo.com