# How do I check for extraneous solutions?

Oct 11, 2014

A lot of times in algebra, especially when you deal with radical functions, you will end up with what you call extraneous solutions. These are solutions to an equation that you will get as a result of your algebra, but are still not correct. It's not that your process is wrong; it's just that this solution does not fit back into the equation (math is very complicating sometimes).

To find whether your solutions are extraneous or not, you need to plug each of them back in to your given equation and see if they work. It's a very annoying process sometimes, but if employed properly can save you much grief on tests or quizzes.

There are two ways you can do this: by hand, and using your graphing calculator. I will go over both, in case you don't have and/or are not allowed to use one on tests/quizzes. In any case, it is helpful to know both so you can get a better feel of how it works, and just to be that much better at math :)

By Hand:

Consider the equation $\sqrt{x + 4} = x - 2$

First, let's solve it using usual algebra:

${\left(\sqrt{x + 4}\right)}^{2} = {\left(x - 2\right)}^{2}$ (square both sides)

$x + 4 = {x}^{2} - 4 x + 4$ (simplify)

${x}^{2} - 5 x = 0$ (subtract $x$ and 4 from both sides)

$x \left(x - 5\right) = 0$. hence $x = 0 \mathmr{and} x = 5$ (zero product property)

Now we have our two solutions; 0 and 5. Now let's plug each of them back into our original equation and see if they work:

5:

$\sqrt{5 + 4} = 5 - 2$
$\sqrt{9} = 3$
$3 = 3$

Therefore, 5 is a verified solution.

0:

$\sqrt{0 + 4} = 0 - 2$
$\sqrt{4} = - 2$
2 ≠ -2

Therefore, 0 is not a solution.

Hence, we would classify 0 as an extraneous solution to this given equation.

By Calculator:

• Set the equation to equal zero
(this ends up being $\sqrt{x + 4} - x + 2 = 0$)
• Plug this into the $y =$ button on your TI-83/84 calculator
• Find the value of each of your solutions (go to 2nd->Calc->Value and enter your solution for $x$)
• You should get zero as an answer for each of them. If you don't, that solution is extraneous.

This is often a much easier way to do it, but as I said above, it is important that you know how to do it by hand as well, as often teachers may ask you to show work, and you may not always have a calculator to help you out.

Hope that helped :)

*equation from http://hotmath.com/*