I was trying to solve this problem:
$$\sqrt{3x+13} = x+ 3$$
So I was pretty confident about this problem and started solving:
$$(\sqrt{3x+13})^2 = (x+ 3)^2$$ $$(3x+13) = (x+ 3)^2$$ $$3x+13 = x^2 + 6x + 9$$ $$0 = x^2 + 3x - 4$$ $$0 = (x+4)(x-1)$$
So my final answer was $x = -4$ and $ x = 1 $. However, it was incorrect because when I plug back in -4 into the original equation I get a extraneous solution. My question is do I always need to plug back in my answers into a radical expression and check if they are valid? Or is there any other way to deduce that there will be a extraneous solution?
$\endgroup$ 24 Answers
$\begingroup$You don't need to plug in values, if you always ensure not to add extraneous solutions.
The equation forces two conditions, namely $3x+13\ge0$ and $x+3\ge0$, which together become $x\ge-3$.
Why $x+3\ge0$? Because $\sqrt{3x+13}\ge0$ by definition (when it exists, of course).
With this condition, you can safely square, because you have an equality between nonnegative numbers. You get (your computations are good) $$ \begin{cases} (x+4)(x-1)=0 \\[4px] x\ge-3 \end{cases} $$ and therefore you know what roots are a solution of the original equation, in this case only $x=1$.
$\endgroup$ $\begingroup$On its domain ($A\ge 0$), note that
$$\sqrt A=B\iff A=B^2\enspace\textbf{and}\enspace B\ge 0,$$
since the symbol$\sqrt{\phantom{h}}$ denotes the non-negative square root of a non-negative real number.
Note that when you squared both sides in the first step, you ended up with the equation
$3x+13 = (x+3)^2$,
which could just as well come from the same original equation but with a negative square root instead:
$-\sqrt{3x+13} = x+3$.
So, at the end of the day, you've solved both equations. To check which solution(s) correspond to which equation(s), you need to plug back in and check.
$\endgroup$ 2 $\begingroup$I think the best or reliable solving is to plot the equations . in your case like below
