Question

Use the equation below to answer these questions.

Answer 1

1) $x+2=\sqrt{3x+10}$

Square both sides to get

$(x+2)^2=(\sqrt{3x+10})^2\implies x^2+4x+4=3x+10\implies x^2+x-6=0$

Factorize and solve:

$x^2+x-6=(x+3)(x-2)=0\implies x=-3,x=2$

2) I assume we take $\sqrt x$ to be defined only for $x\ge0$. Under this condition, a solution would be extraneous if $\sqrt{3x+10}$ is undefined, which happens if $3x+10$.

If $x=-3$, then $3x+10=-9+10=1$, so it is not extraneous.

If $x=2$, then $3x+10=6+10=16$, so it is not extraneous.

So there are not extraneous solutions to this equation.

3) Lots of information on this on the web...