How do you solve #sqrt (3x+10 ) =x# and find any extraneous solutions?

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Answer 1 · 2016-09-02T13:27:52

Solution is #x=5# (#x=-2# is extraneous solution).

Here we can not have #3x+10<0# i.e. #x<-10\3# and any value of #x<-10/3# is extraneous.

Squaring both sides in #sqrt(3x+10)=x#, we get

#3x+10=x^2# or

#x^2-3x-10=0# or

#x^2-5x+2x-10=0# or

#x(x-5)+2(x-5)# or

#(x+2)(x-5)=0# i.e.

either #x+2=0# i.e. #x=-2#

or #x-5=0# i.e. #x=5#

As none of these id less than #-10/3#,

Solution is #x=-2# or #x=5#, but #x=-2# id extraneous as putting #x=-2# in given equation gives us #2=-2#.

Hence solution is #x=5#.