in multiplication and division we change the direction of the inequality symbol. For example: $6>3 \Longrightarrow 6(-2) < 3(-2), \text{ and } 10 > 3 = > \frac{10}{-2} < \frac{3}{(-2)}$.

How does the inequality symbol change direction when $x^2+x-2>0$ ?

The above can be written as: $( x-(-2)\, ) \cdot (x-1) > 0 \Longleftrightarrow ( x+2\, ) \cdot (x-1) > 0$. I am expecting to see:

$x>1$ or $x+2>0 \Longrightarrow x>-2$, but it is wrong.It should be $x<-2$. What am i forgetting to do right to the inequality x>-2? What math property am i missing?

  • $\begingroup$ i fixed that mistake. $\endgroup$ – DontAskTheEye Mar 22 '18 at 15:43

As $( x+2\, ) \cdot (x-1) > 0$ is a product, you need [$x>1$ and $x>-2$] (i.e. $x>1$) OR [$x<1$ and $x<-2$] (i.e. $x< -2$)


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