How do you solve and check for extraneous solutions in #sqrt(23x+3)=14#?

1 Answer
Alan P.
Aug 7, 2015

#x= 193/23# (no extraneous solutions)

Explanation:

Given #sqrt(23x+3) =14#

Square both sides:
#color(white)("XXXX")##23x+3 = 196#

Subtract 3 from both sides
#color(white)("XXXX")##23x=193#

Divide both sides by #23#
#color(white)("XXXX")##x= 193/23##color(white)("XXXX")#approximately 8.391304 according to my calculator

Verify equation with #x=193/23# to ensure this is not extraneous;
#color(white)("XXXX")###sqrt(23(193/23)+3)#

#color(white)("XXXX")##=sqrt(196)#

#color(white)("XXXX")##=14#

So #x=193/23# is a valid result

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