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I'm familiar with extraneous roots. For example $\sqrt{x} = x - 2$

We solve it by squaring both sides \begin{align*} & \implies x = x^2 - 4x + 4\\ & \implies x^2 - 5x + 4 = 0\\ & \implies (x-1) (x-4) = 0\\ & \implies x = 1~\text{or}~x = 4 \end{align*}

But! Only $x = 4$ satisfies the parent equation, $x = 1$ doesn't. Hence $x = 1$ gets rejected.

First I was really surprised, couldn't figure out why $x = 1$ was getting rejected, because $x = 1$ lies in the domain of the parent equation ($\sqrt{x} = x - 2$).

Still can't understand why it's getting rejected. Because when solving equations (or inequalities), we take the intersection between the domain (which is, $x \geq 0$) of the parent equation (or inequality) and the set of solutions obtained on solving the equation (or the inequality). And $x = 1$ here belongs to both, the domain of the equation and the set of $x$ values obtained on solving it. So it's not supposed to get rejected?? Also, is it not supposed to satisfy the equation because as I said, it belongs to both, the domain and the set of values obtained on solving. But it doesn't satisfy. Please explain me what's going on here, why doesn't it satisfy even though it belongs to both, as I've mentioned earlier.

I looked it up, and came to know that such roots are called extraneous roots, and they occur when we raise both sides of a radical equation to an even power. And hence we must always plug the obtained values into the original equation in order to check whether they satisfy the parent equation or not.

My second question is, can extraneous roots occur while solving radical inequalities too? If they can, then how do we get the final solution? Because we have infinite solutions in case of radical inequalities. We can't plug in each and every value to check whether they satisfy or not?

Thanks, and sorry for the long post. I wanted to explain as well as I could, what I don't understand.

N. F. Taussig
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4d_
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3 Answers3

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Every time you transform an equation there is a possibility to introduce extraneous solutions. A very typical case is

$$a=b\to a^2=b^2$$ where $a=-b$ creeps in.

You can either monitor the extraneous solutions as you go transforming, or perform a final rejection.

In your case, moving from

$$\sqrt x=x-2$$ to

$$x=(x-2)^2$$ you can remember that $\sqrt x\ge0$, which implies that $x\ge2$. Then among the options $x=1$, $x=4$, the first one is invalid.

A nastier situation is when a transformation makes you drop some solution, for instance when writing

$$a^2=b^2\to a=b.$$

When this occurs, the final check will not detect the mistake.

  • Thanks a lot. It's starting to make sense now. But when solving complex radical inequalities that involve logarithmic terms too, it's probably not possible to monitor extraneous solutions that creep in as we transform the equation (by squaring it, etc). So I guess I just need to plug the obtained values into the original equation in order to check whether they satisfy or not. What about inequalities? Can extraneous roots creep in while solving inequalities too? I see no reason why they can't. How am I supposed to remove them? We can't plug in each and every values, there are infinite solutions – 4d_ May 21 '18 at 13:49
  • @πtimese: you should remain conscious of the effects of your transformations; mind what I say in the second part. –  May 21 '18 at 13:51
  • What am I supposed to do in case of radical inequalities? I suppose extraneous solutions can creep in when solving radical inequalities too? – 4d_ May 21 '18 at 13:56
  • @πtimese: you need to know the rules of inequality transformations. –  May 21 '18 at 14:13
  • You mean, the rules of inequalities? For example, inequality sign reverses on multiplying both sides by a negative number, or on taking the reciprocal if both sides have the same sign, we shouldn't square unless we know both sides are positive.....and so on. Are you referring to these? Or something else? – 4d_ May 21 '18 at 14:47
  • @πtimese: yes, that's what I mean. When the functions are continuous, the inequalities can change direction at equalities only, so that an alternative method is to solve for the roots and test the signs in between. For example, $\sqrt x>x-2$ will be true for $x<4$. –  May 21 '18 at 14:51
  • I guess we will have two different cases here, when x ≥ 2 and when x < 2. And we solve both the cases separately. That's for clearing up my doubts – 4d_ May 21 '18 at 15:09
  • @πtimese: no, $x<2$ is excluded, we already told you. –  May 21 '18 at 15:10
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From $$(x-1)(x-4)=0$$ we get $x=1$ or $x=4$ since we have $$x\geq 2$$ we get only $x=4$ as the searched solution. For the further questions give an example please!

Dr. Sonnhard Graubner
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  • Sorry, I made typing errors. I just corrected them. But I don't understand why you said x ≥ 2 ?? Isn't it x ≥ 0 ?? If we take √x into consideration? – 4d_ May 21 '18 at 13:34
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    @πtimese: Since $\sqrt{x}$ is never negative, and is supposed to equal $x-2$, it follows that $x-2$ can't be negative either (if $x$ is a solution of the equation). – Hans Lundmark May 21 '18 at 14:04
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In $p$-adics we may define a "principal" square root $\sqrt{1+x}$ as the sum of the Maclaurin series if this converges. Such a Maclaurin series-based definition would reduce to the nonnegative root in the real case (with convergence required in both cases by having $|x|$ sufficiently small), so we may consider this definition a generalization of the usual one.

Then the square root generated by the Maclaurin series in $p$-adics may be the negative one in real analysis, thus impacting the solution set for a radical equation. For an example involving the equation $x=\sqrt{2x+8}$ see here.

Oscar Lanzi
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