How do I find the extraneous solution of sqrt(x+4)=x-2?

1 Answer
Mar 8, 2018

x=0" is an extraneous solution"

Explanation:

"square both sides to 'undo' the radical"

(sqrt(x+4))^2=(x-2)^2

rArrx+4=x^2-4x+4

"rearrange into standard form"

rArrx^2-5x=0larrcolor(blue)"in standard form"

rArrx(x-5)=0

rArrx=0" or "x=5

color(blue)"As a check"

"Substitute these values into the original equation"

x=0tosqrt4=2" and "0-2=-2

2!=-2rArrx=0color(red)" is an extraneous solution"

x=5tosqrt9=3" and "5-2=3larr" True"

rArrx=5color(red)" is the solution"