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This may be standard matter, but how an equation of the form $a x + b + c x^{-1} = 0$ is solved?

scand1sk
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3 Answers3

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$$ax+b+cx^{-1}=0\Leftrightarrow \begin{cases} x\ne 0\\ \frac{ax^2+bx+c}{x}=0\end{cases}\Leftrightarrow\begin{cases} x\ne 0\\ ax^2+bx+c=0\end{cases}$$

and so forth. First step is the same for inequations as well, but for those you would have to evaluate the sign of the whole fraction, not just the numerator.

Added: It might be a matter of debate what happens for $c=0$ and, specifically, if for $c=0$ the quantity $cx^{-1}$ is to be interpreted as the constant $0$ function or as $0$ multiplied by the inverse of $x$: i.e. as if I asked a machine that computes according to PEDMAS to evaluate in $x=0$, $c=0$ and $a,b=\text{something}$ the expression $ax+b+cx^{-1}$, under the additional hypothesis that the machine returns an error as soon as it is asked to evaluate a function outside of its domain. My discussion is in the second case.

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Verify that $x=0$ cannot be a solution (note the negative power). Then multiply throughout by $x$ giving a quadratic.

Deepak
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Assuming $c \neq 0$,

$$a x + b + c x^{-1} = 0 \\ \iff ax^2 + bx + c = 0$$

and solve as you would normally solve a quadratic.

Adam Rubinson
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  • I can't see any good reason to exclude the case $b=0$. –  Dec 10 '20 at 14:24
  • Yeah that's true. There's also not reason to exclude the case $a=0$, right? – Adam Rubinson Dec 10 '20 at 14:27
  • I don't, despite it not being a quadratic anymore, but then again when I have parameters I don't exclude any case. –  Dec 10 '20 at 14:29
  • You mean, you wouldn't exclude the case $c=0$ as well? But OP is probably interested only in $c \neq 0$. – Adam Rubinson Dec 10 '20 at 14:30
  • It's your answer, not mine. :) Personally, I ended up writing about the case $c=0$ than I did of the others. –  Dec 10 '20 at 14:32
  • I can't be bothered to think about the case $c=0$, and it isn't pertinent to OP's question imo. Let OP decide. – Adam Rubinson Dec 10 '20 at 14:43