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Consider these 2 Equations , where multiplication by $x$ is involved.
It is known that multiplication by $x$ should introduce the extraneous solution $x=0$ in both cases.

Observe that the second equation the unwanted $x=0$ solution appears, while that is not occurring in the first equation.

  1. First equation:$$ \begin{split} x+\frac{1}{x} &=2 \quad\implies\text{If I multiply both sides by } x \\ x^2 + 1 &= 2x \\ x^2-2x+1 &=0\\ (x-1)^2&=0\\ x&=1 \end{split} $$
  2. Second equation: $$ \begin{split} 2x+4 &= 0 \quad\implies\text{If I multiply both sides by } x \\ 2x^2+4x & =0\\ 2x(x+2) & =0\\ x=0 \lor x&=-2 \end{split} $$

Why didn't I get $x=0$ as an extraneous solution in the first equation but I did get $x=0$ as an extraneous solution in the second equation?

Daniele Tampieri
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mawaior
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    Because $\frac1x\cdot x=1$ is not true for $x=0.$ – Thomas Andrews Nov 29 '23 at 18:40
  • I don't understand. @ThomasAndrews – mawaior Nov 29 '23 at 18:46
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    Not sure why this got downvoted , +1 from me for new user giving a question which is clear enough , with good effort ! – Prem Nov 29 '23 at 19:09
  • I don't understand why it happens. I multiplied both side by $x$ so I should have got $x=0$ as a solution. (I know that $x$ can't be a solution for the first equation because if $x=0$ than $\frac{1}{x}$ is undefined.) – mawaior Nov 29 '23 at 19:13
  • Here , (1) contains $1/x$ , which implies $x$ can not be $0$ , even when we multiply by $x$. It is not there in (2) , hence when we multiply by $x$ , it could give $0=0$ due to $x$ itself being $0$. That is why we get the Extra Solution. You might check more Examples to clarify. – Prem Nov 29 '23 at 19:14
  • The problem with "multiplying by an expression of $x$" arises when that expression can take the value $0$. So for instance, in the real numbers you could multiply both sides by $x^2+1$ without any issues. In the first case, the original equation has a term $\frac{1}{x}$, which is sort of a general warning that says "and $x$ cannot be equal to $0$" Because of that exclusion, multiplying by $x$ on both sides is a "reversible" operation on your equation, so it does not produce extraneous solutions. That does not occur in the second equation. – Arturo Magidin Nov 29 '23 at 19:25
  • If your second equation was "$2x+4=0$ and $x\neq 0$", then multiplying by $x$ would lead to "$2x(x+2)=0$ and $x\neq 0$", which would then result in $x=-2$ alone (since $x=0$ is explicitly disallowed), so you would also not have an extraneous solution in the second equation. The first equation has that "and $x\neq 0$" implicit in its form. – Arturo Magidin Nov 29 '23 at 19:28
  • $x=y \Rightarrow f(x)=f(y)$ but $f(x)=f(y) \nRightarrow x=y$. (If $x=a$ is a solution for the original equation then it is also a solution for the $f(x)=f(y)$ but not necessarily vice versa)

    Is that the reason we get extraneous solutions? @ArturoMagidin

     – mawaior Nov 29 '23 at 19:32
  • It's a reason you may get extraneous solutions. – Arturo Magidin Nov 29 '23 at 19:34
  • Oh thank you, I think I understand. So we will never get extraneous solutions when multiplying by any expression in the denominator, right? @ArturoMagidin – mawaior Nov 29 '23 at 19:35
  • It is the only reason that we may get extraneous solutions, right? @ArturoMagidin – mawaior Nov 29 '23 at 19:40
  • We may get extraneous solutions when $f(a)=f(b)$ has solutions even if $a\neq b$; that is, when the function $f$ is not one-to-one. Here, $f$ is "multiply by $x$" which is one to one when $x\neq 0$, but not one to one when $x=0$. However, if it just so happens that the cases where $f(a)=f(b)$ holds even with $a\neq b$ are solutions to the original equation, then you won't get any. If you start with $2x=x$, then multiply by $x$ and deduce $x=0$, the fact that multiplication by $x$ may introduce $x=0$ as an extraneous solutions doesn't actually do so, as it was already a solution. – Arturo Magidin Nov 29 '23 at 19:52

1 Answers1

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As you note in the comments, part of the issue is that when you perform some kind of transformation $f$ to the equation, you really want that $$a=b \iff f(a)=f(b)$$ hold; for arbitrary expression, we only get the $\implies$ direction, so that a solution to $f(a)=f(b)$ need not yield a solution to $a=b$.

But whether the $\Longleftarrow$ implication holds may depend on the domain of interest: what kind of values $x$ is allowed to take. (Because restricting the values may turn a function that is not invertible/one-to-one into one that is). For example, if you were working exclusively in positive real numbers, then you would get no solutions at all in the second equation, either before or after you multiply by $x$. If you are only interested in real solutions, you can multiply both sides of an equation by $x^2+1$ and that will never produce extraneous solutions, because $x^2+1$ is never equal to $0$. But if you were looking for solution in $\mathbb{C}$, that expression would introduce potential extraneous solutions.

I'm assuming you are working in $\mathbb{R}$, the real numbers.

When you multiply by an expression involving $x$, to ensure that you do not change the solutions you need to make sure that expression is not equal to $0$. The way to keep track of this is to either consider the case where the expression is equal to $0$ separately, or check if that case happens to solve the equation.

So for example to ensure that multiplication by $x$ is reversible, you should add the clause "and $x\neq 0$" to your equations, and then consider what happens if $x=0$. Likewise if you divide by, say, $x-1$ (you would be saying "and $x\neq 1$", and need to consider the case $x=1$ separately).

Look at your first equation, $$ x+\frac{1}{x} = 2.$$ There is already an implicity "$x\neq 0$" in that equation, because the expression on the left is not defined at $x=0$. It's like having a function $f(x)$ defined by $f(x) = x+\frac{1}{x}$, and you want to figure when it takes the value $0$. Because $0$ is not in the domain of $x$, you will never take $x=0$: it's disallowed.

So what you have, with everything explicit, is: $$\begin{align*} x+\frac{1}{x}&=2&&\text{and }x\neq 0\text{ because of the }\frac{1}{x}\\ x\left(x+\frac{1}{x}\right) &= x(2) &&\text{and }x\neq 0\text{ because of the }\frac{1}{x}\\ &&&\quad\text{ and because we multiplied through by }x\\ x^2+1 &= 2x &&\text{and }x\neq 0\text{ for multiple reasons.} \end{align*}$$ So you didn't change the valid domain of your computations when you multiplied everything by $x$, because you had already excluded it when you wrote down your initial equation.

By contrast, in your second equation you do not have that initial exclusion, so when you multiply by $x$ you do change the scope: $$\begin{align*} 2x+4&= 0\\ x(2x+4) &= x(0) &\text{and }x\neq 0\text{ because we multiplied through by }x\\ x(2x+4) &= 0 &\text{and }x\neq 0\text{ because we multiplied through by }x \end{align*}$$ Solving this will give only $x=-2$ (because $x=0$ is explicitly excluded). To be thorough, one would now check to see if $x=0$ happens to be a solution to the original equation (because we excluded it for no good reason during our computations). By contrast, in the first equation we don't need to check whether $x=0$ happens to be a solution because it was excluded a priori, and not by our manipulation of the expression.

Arturo Magidin
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  • Thank you for you explanation!

    What about $f(x)=\sqrt{x}$? $x=y\implies \sqrt{x}=\sqrt{y}$ but if I have $x^2=9 \implies \sqrt{x^2}=\sqrt{9} \quad x=3$ but $x=3$ isn't the only solution for $x^2=9$. Isn't that a contradiction to $x=y \implies f(x)=f(y)$?

     – mawaior Nov 29 '23 at 20:40
  • @mawaior Actually, $\sqrt{x}=\sqrt{y}\implies x=y$, because the square root is only defined for nonnegative arguments. But $x=y$ does not imply $\sqrt{x}=\sqrt{y}$, because the latter may not be defined. Your error above is that $\sqrt{x^2}=|x|$, not $x$. So $x^2=9$ gives $\sqrt{x^2}=\sqrt{9}$, which gives $|x|=3$, which gives $x=3$ or $x=-3$. – Arturo Magidin Nov 29 '23 at 22:01