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

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Jim G.
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Mar 8, 2018

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#"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"#

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