I have made a rather obvious yet peculiar observation while calculating with quadratic inequalities. Take a simple quadratic inequality like the one below


by multiplying both sides by $x$, then subtracting $x$ from both sides we get


Hence, both the above inequalities are one and the same in theory. However, the solution to the first inequality is $x>0$, while the second inequality is satisfied for all $x\in\mathbb R$.

It is not difficult to explain this simple case, seeing that the LHS of the first inequality will turn negative for negative $x$, but my question is more of a general nature, if you like. I wonder, why do two identical (but paraphrased/rewritten) inequalities have different solutions?


If you multiply an inequality with a negative number, you have to exchange the inequality signs $<$ and $>$.

For instance, multiplying $1<2$ with $(-1)$ you get $$-1>-2.$$

  • $\begingroup$ That is correct. I didn't multiply it with a negative number though.. I'm afraid this doesn't explain the different solutions either. $\endgroup$ – Milo Wielondek Apr 14 '11 at 16:11
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    $\begingroup$ Well, $x\in\mathbb R$ is negative if $x<0$, right? So, unless $x>0$ you are not allowed to multiply your inequality with $x$ without exchanging the inequality signs. $\endgroup$ – Rasmus Apr 14 '11 at 16:12
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    $\begingroup$ Ok, I see where you're going! This would mean that I cannot rewrite that inequality at all for $x\in\mathbb R$ since it both means I would have to flip the inequality sign (for $x<0$) and leave it untouched ($x>0$)? What is a convenient way of solving such inequalities then, without rewriting them? $\endgroup$ – Milo Wielondek Apr 14 '11 at 16:13
  • $\begingroup$ I guess that a case differentiation is in order then. $\endgroup$ – Rasmus Apr 14 '11 at 16:21
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    $\begingroup$ It just means that you treat the two cases $x<0$ and $x>0$ separately. Then, in each case, you know what the sign of $x$ is. $\endgroup$ – Rasmus Apr 14 '11 at 16:29

the method i describe below helps finding a complete solution by sticking to the simple principle that for polynomials and rational functions only candidates for the sign change is across the zeros of the numerator and denominator.

instead of looking at $\frac{x^2 + 1}{x} > 1$ let us look at $\frac{x^2 + 1}{x} -1$ which simplifies to the rational function $\frac{x^2 - x + 1}{x}$. the numerator has no real roots so it is of one sign which is positive because the value at 0 is 1. sign of rational function is now the sign of numerator.


Another way of thinking about of some of the previous answers and comments is to use a geometric point of view. Draw a graph of the function y = f(x) =((x^2+ 1)/x) - 1. Often this can be done without specialized algebraic knowledge. When the graph is above the line y = 0 you can read off the answer from what you have graphed.


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