Why can't I take the square root of both sides when finding the roots of a quadratic equation?

Question

Is $x^2 - x - 12 = 0$ equivalent to $x = \sqrt{x + 12}$?

I started with

$x^2 - x - 12 = 0$

and made the following changes:

$x^2 - x - 12 = 0$
$x^2 = x + 12$
$x = \sqrt{x + 12}$

From here I can eyeball it and see that x = 4 and x = -3 are solutions.

I know there is a better way to find the roots, but I was told that $x^2 - x - 12 = 0$ and $x = \sqrt{x + 12}$ are not equivalent. If they are not, why not?

What about $x = -3$? Do you get that from $x = \sqrt{x+12}$? — Daniel Fischer, Sep 15 '13 at 23:55
If you're going to "eyeball it" you might as well just guess $4,-3$ from the initial equation. — vadim123, Sep 16 '13 at 00:11
I found this question helpful too: http://math.stackexchange.com/questions/61252/is-there-a-name-for-this-strange-solution-to-a-quadratic-equation-involving-a-sq?rq=1 — dumb question, Sep 16 '13 at 00:21
$x = \pm \sqrt{x + 12}$ would be the correct way to express it. — Steven Alexis Gregory, Sep 28 '16 at 22:08

score 6 · Accepted Answer · answered Sep 15 '13 at 23:56

It's not equivalent for two reasons:

We don't know that $x + 12$ is non-negative, so it might not be valid to take square roots
$\sqrt{x + 12}$ is always non-negative by definition, provided it's defined. So $x = -3$ is not actually a solution to this new equation.

Why can't I take the square root of both sides when finding the roots of a quadratic equation?

1 Answers1