2
$\begingroup$

We have extraneous roots when we perform operations on the given equation which are not invertible for all or some values of the variable in that equation.

For example, if we have $x+2=0$, then, multiplying both sides by zero may give us $0=0$. It is not invertible since division by zero doesn't make sense.

Solving an equation in one variable is the process of converting equations into equivalent equations and equivalent equations into equivalent equations and doing it so until the solution is obvious.

Although the above equation $x+2=0$ can painlessly be converted into $x=-2$ and solution would be clear but ignoring it and multiplying it by $x$ would convert it into an equation which would be invertible and that equation is, $x^2+2x=0$

We get solutions $-2$ and $0$ but $0$ is an extraneous root as $2 \neq 0$.

Of course, this is not the only related case but there may be infinite.

In the case above, we multiplied by $x$ and this took us to an extraneous root. So it is clear that operations performed may lead us to extraneous root.

My question is; are there signs which could tell us that performing this and that operation on an equation may leave us with extraneous root(s)? Is there any method to predict that we are going to get extraneous roots or all the time we have to check it by substitution?

$\endgroup$
  • $\begingroup$ See: math.stackexchange.com/questions/55445/… $\endgroup$ – Harto Saarinen Dec 8 '15 at 21:37
  • $\begingroup$ As you said the process is to go through a string of equivalent equations until we have one we're happy with. Extraneous roots occur when your equations aren't logically equivalent. $x+2=0 \Rightarrow x^2+2x=0$, but $x^2+2x=0 \nRightarrow x+2=0$ since dividing by $x$ is not allowed when $x=0$. $\endgroup$ – rVitale Dec 8 '15 at 21:39
  • $\begingroup$ I know that equations are not equivalent. I didn't said that I am converting $x+2=0$ into an equivalent equation by multiplying it by $x$. The reason of telling about this process was to support the statement just next to it. However it hardly contributes there. $\endgroup$ – Sufyan Naeem Dec 8 '15 at 21:45
  • 1
    $\begingroup$ @HartoSaarinen, unfortunately the question you shared, is already answered in the very first para of my post. :) $\endgroup$ – Sufyan Naeem Dec 8 '15 at 21:59
1
$\begingroup$

A common case (e.g. often encountered in equations involving square roots) is taking an even power of both sides of the equation.

$\endgroup$
0
$\begingroup$

Multiplying by $0$ is the "sign", because it goes only one way (and you can't "go back"). So when you multiply by $x$, you simply must make sure you can exclude the case $x=0$.

In algebraic words $A=B\Leftrightarrow f(A)=f(B)$ is true if and only if $f$ is invertible. (Multiplying by $0$ is not invertible.)

$\endgroup$
0
$\begingroup$

Yes you can. Whenever you change the degree of equation while performing operations, you may get extraneous roots. So whenever you do this, check the solutions by actual substitutions. When you not make such operations, just write down the solution in the solution set without any worry.

$\endgroup$
0
$\begingroup$

Suppose that somewhere along the way in solving an equation, you need to multiply both sides by $(x-2)$, for example to clear the denominator of a fraction.

This can only give you an extraneous root if $x = 2$ is one of your possible solutions at the end, since otherwise multiplying by $(x-2)$ is an invertible operation.

How can you predict if $x = 2$? You could plug $x = 2$ into both sides and, if there's no other variables or at least none that get in the way too much, see if you get something obviously true (like $2 = 2$), or something obviously false (like $1 = 2$) in which case you didn't get an extraneous root. But this is often as much effort as just solving the equation in the first place.

Of course if you did get $x = 2$ as a possible solution, you also need to plug $x = 2$ back into the original equation to see if it's a genuine root or an extraneous one.


Next, consider $\sqrt{x - 2} = 3$. Squaring both sides is a non-invertible operation, at which point you have $x - 2 = \pm 9$ so $x$ could be $11$ or $-7$, but if you're working over the reals then $\sqrt{(-7)+2}$ isn't defined so this root is extraneous.

All you can predict at the point when you square is that any solutions where $x < 2$ will be extraneous. Whether or not there is such a solution depends on the form of the equation and the number on the right: the equation $\sqrt{x - 2} = c$ has two potential solutions $c^2 + 2$ (always positive, therefore genuine) and $- c^2 + 2$ which is genuine for $c \leq \sqrt 2$ and extraneous otherwise. To work this much out, you have to solve the equation in the first place.

If you imagine that both the term under the root and the term on the right can be arbitrary terms in $x$, you can imagine there's no general way of predicting what roots you'll get - if after squaring you're left with a quadratic equation, you have to solve a quadratic equation just to find the potential roots.


What you can and should do at the point where you apply a non-invertible operation is work out what potential extraneous roots this operation could introduce, make a note of them and then check all the roots you get at the end. I'm afraid there's no general-purpose short cut.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.