How do you solve #sqrt(9x+10)=x# and find any extraneous solutions?

1 Answer
May 27, 2016

Answer:

#x=10# or #x=-1#
(no extraneous solutions)

Explanation:

Given
#color(white)("XXX")sqrt(9x+10)=x#

Square both sides (this where an extraneous solution might be introduced)
#color(white)("XXX")9x+10=x^2#
Rearrange as a quadratic in standard form
#color(white)("XXX")x^2-9x-10=0#
Factor
#color(white)("XXX")(x-10)(x+1)=0#
Either
#color(white)("XXX")(x-10)=0color(white)("XX")rarrcolor(white)("XX")x=10#
or
#color(white)("XXX")(x+1)=0color(white)("XX")rarrcolor(white)("XX")x=-1#

Checking against original equation:
#color(white)("XXX")#If #x=10#
#color(white)("XXXXXXX")sqrt(9x+10)=sqrt(9(10)+10)=sqrt(100)=10=x#
#color(white)("XXX")#This solution is valid.

#color(white)("XXX")#If # x=-1#
#color(white)("XXXXXX")sqrt(9x+10)=sqrt(9(-1)+10)=sqrt(1)=1=x#
#color(white)("XXX")#This solution is valid