How to determine which equality sign to use for $x(2x+7) > 0$ Express $x(2x+7)>0$ in interval notation. How do you determine which sign to use for the answer? Here is what I did,
$2x+7>0$ and $x>0$. I understand $x>0$ but why is $x<-\dfrac{2}{7}$? Shouldn't the sign be greater than because I am dividing by $2$ and not $-2$? Or does the $-7$ affect the sign as well?
 A: There are two cases, because $ab>0$ if and only if $a>0$ and $b>0$, or $a<0$ and $b<0$. If $x>0$, then automatically $2x+7>0$, so $x(2x+7)>0$, and you have a solution. However, there are also solutions when $x<0$ and $2x+7<0$. Here the stronger constraint is the second one: if $2x+7<0$, then $2x<-7$, and $x<-\frac72$ (not $-\frac27$), in which case it’s certainly true that $x<0$ and hence that $x(2x+7)>0$.
In other words, every $x>0$ is a solution, but so is every $x<-\frac72$.
A: Let $A=2x+7$ and $B=x$. then we have $AB>0$ what are the conditions which this inequality held? the clear answer (using the real numbers properties) is when $A$ and $B$ are (both) positive, or $A$ and $B$ are (both) negative. in brief, we conclude as follows:
1- $A>0\;and\;B>0$
2- $A<0\;and\;B<0$
thus
$$A=2x+7>0\quad and\quad B=x>0$$
$$A=2x+7<0\quad and\quad B=x<0$$
therefore
$$x>-\frac{7}{2}\quad and\quad x>0$$
$$x<-\frac{7}{2}\quad and\quad x<0$$
Note that these terms , shows two sets: first is $$S=\{x\in\mathbb{R}\;;\;x>-\frac{7}{2}\quad and\quad x>0\}=\{x\in\mathbb{R}\;;\; x>0\}$$
and second is $$T=\{x\in\mathbb{R}\;;\;x<-\frac{7}{2}\quad and\quad x<0\}=\{x\in\mathbb{R}\;;\;x<-\frac{7}{2}\}$$
Notice the word or we mentioned it before, we conclude the answer with $$S\cup T=\{x\in\mathbb{R}\;;\;x<-\frac{7}{2}\quad or\quad x>0\}$$
You may draw the real line to find out the details.
A: Product of two numbers $ab>0$ if and only if either $a<0, b<0$ or $a>0, b>0$.
