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I am currently in a Pre Calculus class at my High School. I have come across the concept of extraneous solutions, particularly when solving absolute value equations, radical equations, and logarithmic equations. My question is, why do these solutions exist?

My teacher never explained this, which is understandable given that I am in a High School math class, and there isn't much time for the teacher to go into the actual derivations of everything. I am wondering because I plan to major in Mathematics, and having a conceptual understanding of this is important to me.

If anyone could explain the reason extraneous solutions exist for the three examples I noted, it would be of great help to me.

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One reason extraneous solutions exist is because some functions are not injective. In particular, given an equality

$$a(x)=b(x)$$

we can square both sides to get

$a(x)^2 = b(x)^2$. It is true that $a(x)=b(x) \Longrightarrow a(x)^2 = b(x)^2$; however, it is not true that $a(x)^2 = b(x)^2 \Longrightarrow a(x)=b(x)$. In particular, we may have $a(x) = -b(x)$.

Consequently, we may have solutions $x$ to the equation $a(x)^2 = b(x)^2$ which satisfy $a(x) = -b(x)$ but not $a(x) = b(x)$. These are extraneous. In general, you will not lose solutions squaring both sides, but gain extraneous ones.

Another reason is domain considerations. Consider the logarithmic equation

$\log(f(x)) + \log(g(x)) = \log(h(x))$

We thus solve $f(x) \cdot g(x) = h(x)$. We might get a bunch of solutions $x$ to this equation, but they are only valid so long as $f(x)$ and $g(x)$ are greater than zero, because the identity $\log(x) + \log(y) = \log(xy)$ only works for positive $x,y$. However, again, you will not lose solutions with this method. If $x$ satisfies $\log(f(x)) + \log(g(x)) = \log(h(x))$, then that $x$ also must satisfy $f(x) \cdot g(x) = h(x)$.

You have to remember what you do when you solve an equation. It is easy to forget since it is almost second nature to us. What you do is alter the given equation such that solutions to the equation are preserved. When adding and subtracting, there is never a problem, since adding and subtracting are bijective maps. Other functions may not be as nice, and your altered equation may yield solutions which do not solve your original equation.

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Those "solutions" appear because in the process of solving the equation you do things which are not reversible.

For example, suppose you want to solve the (easy!) equation $$x=-1.$$ If you square both sides, you obtain $$x^2=1,$$ and this can be solved using the formula for roots of a quadratic equation (!) to find the two solutions $1$ and $-1$. Only one of these is a solution of the original equation.

Can you spot the non-reversible step I took?

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  • $\begingroup$ Im confused by what you mean when you say non-reversible. $\endgroup$ – Nick Mar 28 '16 at 5:07
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    $\begingroup$ Well, non-reversible means simply that cannot be reversed. In the example: if a number $x$ is equal to $-1$, then clearly the square of $x$ is equal to the square of $-1$, namely, to $1$. Therefore the operation of squaring both sides of the equality «$x=-1$» produces a new equality «$x^2=(-1)^2$» which is a consequence of the first one. You cannot undo this: if you have a number $y$ such that $y^2=(-1)^2$, then you cannot remove the squares to obtain the equation $y=-1$. $\endgroup$ – Mariano Suárez-Álvarez Mar 28 '16 at 5:10
  • $\begingroup$ I was under the impression that I could solve any equation $a^{n} = b^{n}$ by taking the $n$th root of both sides. Of course if $n$ was even, that would yield me two different answers, one negative and one positive. Wouldn't that be reversing the operation? Maybe I am misguided, but I just don't see where extraneous solutions fit in. $\endgroup$ – Nick Mar 28 '16 at 5:15
  • $\begingroup$ I suggest you talk to your instructor. $\endgroup$ – Mariano Suárez-Álvarez Mar 28 '16 at 5:32
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The reason extraneous solutions exist is because some operations produce 'extra' answers, and sometimes, these operations are a part of the path to solving the problem.

When we get these 'extra' answers, they usually don't work when we try to plug them back into the original problem.

Squaring is a common operation that produces multiple values. As Mariano Suárez-Alvarez♦ notes,

$$x=-1\tag1$$

$$\implies x^2=(-1)^2=1$$

$$\implies x^2-1=0$$

$$\implies x=\pm1$$

But obviously, we see $x\ne+1$, as you can see by $(1)$.

Reading your comments, I see you present an example:

$$a^n=b^n\implies a=b$$

This is only partially true, for the full algebraic solution to this problem is given as

$$ae^{\frac{2}ni\pi x}=b$$

Where $x=0,1,2,3,\dots$

For $x=0$, this breaks down into $ae^0=a=b$, but that is not the full picture.

So if we have something like $x=a\implies x^n=a^n$, the latter equality produces many different results, whereas the original equality has only one result.

Extraneous solutions for general problems like $x+a=\sqrt{x+b}$ are actually quite interesting, but further understanding of things like branches and complex numbers must be understood to fully grasp the meaning of those extraneous solutions. (If you really want to, solve some of these square root problems using the quadratic formula, and note which of the two solutions the quadratic formula game came out right ($+$ or $-$?))

Lastly, extraneous solutions when dealing with logarithms are simply due to your lack of understanding of how complex numbers play into logarithms. When you must use the definition that a logarithm is only defined for positive real input, then you will get extraneous solutions for the very reason that you have that parameter in place. Once you learn how to deal with complex logarithms, I don't believe you can have extraneous solutions in logarithms.

I can't remember doing any extraneous solutions for absolute values, but I'm sure there is an explanation for those.

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  • $\begingroup$ Perhaps we are meant to infer that taking absolute values can introduce extraneous roots. E.g. $x=-1$ implies $|x|=1$ but the converse is false. $\endgroup$ – hardmath Apr 20 '16 at 5:53
  • $\begingroup$ @hardmath Ah, yes, that it is quite right. $\endgroup$ – Simply Beautiful Art Apr 20 '16 at 16:59
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One source of extraneous solutions is from definitions. If you want to solve $$x+a=\sqrt{x+b},$$ usually you square both sides and solve the resulting equation: $$(x+a)^2=x+b.$$ However, the square root is defined to be nonnegative, whereas there might be some solutions where $x+a<0$.

EXTRA:

As for logarithms, it is possible for $\log(xy)$ to be defined without $\log(x)+\log(y)$ being defined; the combined form allows for more possibilities ($x,y<0$ or $x,y>0$) than the expanded form ($x,y>0$).

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In general, in mathematics and the real world, you get EXTRANEOUS ROOTS any time you are initially presented with (via some mechanical / automated / canonical process) a SUPERSET of the set that you want, and the sifting out of that desired set from the superset is left to you, for example, when panning for gold, or when reading the owner’s manual for your vehicle: “This owner’s manual covers all models of your vehicle. You may find descriptions of equipment and features that are not on your particular model.” (Getting rid of square roots by squaring is a well known instance of such a canonical process.)

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