I posted on another forum but haven't got a reply:

If we followed the equation

25m^2+ 20m>0
.: 5m(5m +4) >0

m>0, but wouldn't m be equal to m>-4/5 not m<-4/5 since we aren't multiplying or dividing both sides by a negative number?

Alot of thanks if anybody can help me out

2. The method I learnt was to actually draw a graph of the equation, and mark the roots. You need where the graph is positive, and you can easily see from the graph I've drawn that the set of points satisfying y>0 are:

$\lbrace{m:m<-\frac{4}{5}\rbrace} \cup \lbrace{m: m > 0 \rbrace}$

3. Originally Posted by delicate_tears
I posted on another forum but haven't got a reply:

If we followed the equation

25m^2+ 20m>0
.: 5m(5m +4) >0

m>0, but wouldn't m be equal to m>-4/5 not m<-4/5 since we aren't multiplying or dividing both sides by a negative number?

Alot of thanks if anybody can help me out
Factoring an inequality is quite insufficient.

If x*y = 0, then we can say x = 0 or y = 0.

If x*y > 0, there is not so simple a statement.

If x*y > 0, then (x > 0 AND y > 0) or (x < 0 AND y < 0)

You must find where BOTH conditions are met simultaneously.

4. Originally Posted by delicate_tears
25m^2+ 20m>0
The standard method (if there is one) to do these is the following:
$25m^2 + 20m > 0$

Make sure everything is on one side of the equation. (Already done in this problem.)

Now find out what values of m make the LHS 0 ("critical points.")
$5m(5m + 4) = 0$

So $m = -\frac{4}{5}$ and $m = 0$.

Now break the real number line into pieces based on these critical points:
$\left ( -\infty, -\frac{4}{5} \right ), \left ( -\frac{4}{5} , 0 \right ), (0, \infty )$

Now test the inequality on each one of these intervals:
$\left ( -\infty, -\frac{4}{5} \right )$: $5m(5m + 4) > 0$

$\left ( -\frac{4}{5} , 0 \right )$: $5m(5m + 4) < 0$

$(0, \infty )$: $5m(5m + 4) > 0$

So it looks like the solution set is
$\left ( -\infty, -\frac{4}{5} \right ) \cup (0, \infty )$.

-Dan