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"#