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

1 Answer
Alan P.
May 27, 2016

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

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