How do you solve #x= sqrt(3x+4)# and find any extraneous solutions?

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Answer 1 · 2017-04-17T13:21:21

We first conclude that #3x+4>=0->x>=-4/3#
The result of a radical may not be negative, so #x>=0#

Now we can square both sides:

#->x^2=3x+4#

Putting everything to one side we get:

#->x^2-3x-4=0#

Factoring:

#->(x+1)(x-4)=0->x=-1orx=+4#

After the initial conditions only #x=4# is left.