# How to know if there is a extraneous solution in a radical expression

I was trying to solve this problem:

$$\sqrt{3x+13} = x+ 3$$

$$(\sqrt{3x+13})^2 = (x+ 3)^2$$ $$(3x+13) = (x+ 3)^2$$ $$3x+13 = x^2 + 6x + 9$$ $$0 = x^2 + 3x - 4$$ $$0 = (x+4)(x-1)$$

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?

• In this case you can tell there is only one solution simply by thinking about the graphs of the functions on the left and right sides of your original equation. This tells you one of your roots is extraneous. It is also clear graphically that that solution is positive. So the extraneous root must be $-4$. Alternatively, recall that the range of the square root function is the set of nonnegative reals. So your original equation implies $x\geq-3$. – symplectomorphic Feb 18 '17 at 22:58
• @Khosrotash I am trying to avoid checking the final answers because it is a timed test and I need to move really quickly – Pablo Feb 18 '17 at 22:59

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 $x=1$.

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.

• Amm so does that mean that every time I square a radical then I would get something like this $$±(3x+13)^2$$ – Pablo Feb 18 '17 at 23:01
• No, because the square of a number is always positive. It just means that when you square both sides to try and solve the original equation, you're actually solving two equations at once. – J Richey Feb 18 '17 at 23:02

I think the best or reliable solving is to plot the equations . in your case like below