How to properly divide equations and inequalities by a variable?

Question

Let's consider a simple equation:

$$3x^2-2x+x=x^2-1$$

or inequality

$$3x^2-2x+x < x^2-1$$

or

$$3x^2-2x+x > x^2-1$$

How to properly divide by $x^2-1$. In many examples I've seen it is stated that division can be done properly only if $x^2-1$ is squared so it is always greater than zero, but no deeper explanation is given. Could somebody give wide explanation of this?

Thanks in advance

Well, you can't divide by zero, so you need $x^2 \neq 1$. If your divide the inequalities, you need to take account of the sign of $x^2-1$ (if negative, it 'flips' the inequality). — copper.hat, Sep 25 '13 at 06:09

score 4 · Accepted Answer · answered Sep 25 '13 at 06:10

4

If the expression by which you are dividing is smaller than zero, then the inequality reverses.

Example : $-3<2$, dividing both sides by $-1$, we have the inequality reversed or $3>-2$.

Therefore you should check cases for if $x^2-1<0$ then after dividing reverse the inequality or if it is greater than zero then inequality does not get reversed.

answered Sep 25 '13 at 06:10

Shobhit

6,902

So I should consider two (or three) parts of the domain? One for $x^2-1>0$ and two for $x^2-1<0$ ? – Misery Sep 25 '13 at 06:34
1

yes. $1)$ $x^2-1<0$ and $2)$ $x^2-1>0$ and $3)$ $x^2-1 \neq 0$. – Shobhit Sep 25 '13 at 06:37
And what with equality? – Misery Sep 25 '13 at 06:55
1

There is no need to consider cases in equality simply take the terms in R.H.S to L.H.S, simplify, and apply splitting middle term or quadratic formula – Shobhit Sep 25 '13 at 07:01

score 2 · Answer 2 · answered Sep 25 '13 at 08:26

2

There in no need to deal with cases. Assume, $f$ and $g$ are continous function, defined on some interval. To solve $f(x)<g(x)$, solve $f(x)-g(x)=0$. Between consecutive zeros the sign of $f-g$ is constant.

Michael

answered Sep 25 '13 at 08:26

Michael Hoppe

18,103
3
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How to properly divide equations and inequalities by a variable?

2 Answers2