basic inequality using quadratic equation

32 Views Asked by At

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?

1

There are 1 best solutions below

0
On BEST ANSWER

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. $x< -2$)