Question 173483
{{{sqrt(x+10)+2=x}}}
:
{{{sqrt(x+10)=x-2}}}
:
{{{x+10=(x-2)^2}}}squaring both sides
:
{{{x+10=x^2-4x+4}}}
:
{{{x^2-5x-6=0}}}
: 
 it works in the quad equation but the only way it works in the original is if the original left side had a negative in front of the radical{{{-sqrt(x+10)}}}
:
When solving radical equations, extra solutions may come up when you raise both sides to an even power.  These extra solutions are called extraneous solutions. If a value is an extraneous solution, it is not a solution to the original problem. 
In radical equations, you check for extraneous solutions by plugging in the values you found back into the original problem. If the left side does not equal the right side, then you have an extraneous solution. 

:therefore -1 is not a solution to this equation
:
{{{highlight(x=6)}}}

*[invoke quadratic "x", 1, -5, -6]