How do I find the extraneous solution of sqrt(x+4)=x-2?
1 Answer
Mar 8, 2018
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"