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.

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