# Extraneous Roots

When dealing with extraneous roots and checking them, if you find that one of the, two roots let's say, is extraneous, then does this imply that the other root is real or would one need to check the other root as well...so potentially it has no roots?

I'm pretty sure the one root being extraneous would imply that the other root is real just by thinking of it logically but I cannot ever remember being told this and cannot find verification anywhere else.

• What do you mean by "extraneous"? Do you mean that like complex roots or real roots? And do you mean to only ask about quadratic polynomials, or about polynomials of all degrees? – Milo Brandt Nov 5 '15 at 2:23
• I probably shouldn't have used the word real...I mean roots of an equation that are not true solutions. Like in high school algebra; if you square both sides of an equation, you may end up with a value of x that is not a true solution. The example I gave and the problem I'm looking at is a quadratic equation yes, but I'd imagine the same logic would apply to a polynomial of arbitrary degree – Liam Cooney Nov 5 '15 at 2:25
• Ah; in that case, if you know there is an answer (i.e. if I throw this ball in the air, it's definitely coming back down eventually), then if all but one of the roots is nonsense, the remaining root had better make sense. (If it doesn't, that's a good indication that you've not done something right) – Milo Brandt Nov 5 '15 at 2:36
• There is no physical context in the problem, so I don't KNOW there is an answer. Nonetheless, I thought that if you had two roots and one was extraneous then the other must not be, but it appears that I am wrong (see answer below). Thank you! – Liam Cooney Nov 5 '15 at 2:38
• If you have you have $\forall x (F(x)=0 \to G(x)=0)$ then you have $\{x :F(x)=0\}\subset \{x: G(x)=0\}.$ This is not enough to infer that $\exists x (F(x)=0)$. – DanielWainfleet Nov 5 '15 at 3:08

There are (at least) two ways to interpret "extraneous roots".

One is that roots relate back to a physical context. In such a scenario, it is not necessarily the case that there are any "real" roots at all—real here referring not to whether or not it's a complex number (involving $i$, the square root of $-1$), but whether it fits the physical context. Often, it's merely negative, and because the variable is modeling a non-negative number (e.g., the number of books in a pile), the root is considered extraneous.

If, say, a quadratic equation has two roots, and neither of them fits the physical context, then both of them are extraneous, and there are no "real" solutions at all. That just means that the constraints of the problem cannot be satisfied.

Another way to interpret "extraneous roots" refers specifically to roots developed because of a non-invertible transformation of some original equation. (Squaring is the most common example of this transformation, as you point out.) Under what circumstances this happens and what its implications are is a different set of questions.

There still isn't guaranteed to be a non-extraneous root, though. Consider what happens when you square both sides of

$$-|x| = \sqrt{2x^2-1}$$

One obtains

$$x^2 = 2x^2-1$$

or

$$x^2-1 = 0$$

which clearly has the roots $x = \pm 1$, but neither of those is a root of the original equation.

• And thus it is necessary to check both. Thank you! – Liam Cooney Nov 5 '15 at 2:28

Extraneous roots occur when you apply a function to both sides of an equation that is not 1-1. Here is a very trivial example. Consider the equation $$x = 1.$$ Square both sides and we get $$x^2 = 1,$$ which yields solutions $x = \pm 1$. The second, extraneous, root occurred because squaring destroys information about the sign of a real number.

Suppose the equation to solve is $$x^2=-1$$ and you square both sides: $$x^4 = 1$$ Now the resulting equation has two real roots, $x=-1$ and $x=1$, none of which is a root of the first equation.

So the answer is: YES, of course you should check ALL roots for validity (unless you proceed carefully, so that you can prove that no step of your solution introduces extraneous roots, in which case there simply will be no extra roots).