Question

How do you solve `2x^2+3x>0`?

Answer 1

`2x^2+3x>0 <=> x in (-oo, -3/2)uu(0,oo)`

Explanation:

The trick is to note that the sign of a polynomial expression can only be different at two points if the expression evaluates to `0` at some point between them. Then, if we consider the intervals on either side of where the expression evaluates to `0`, we can find all points where it is positive

Solving for where it is `0`:

`2x^2+3x=x(2x+3)=0`

`<=> x=0 or 2x+3=0`

`<=> x=0 or x= -3/2`

Thus the intervals in question are `(-oo, -3/2), (-3/2, 0), (0, oo)`

From here, we can just test a point in each to see what the sign of the expression will be:

`2(-2)^2+3(-2)=2(4)-6=8-6=2>0`

Thus `2x^2+3x>0` on `(-oo,-3/2)`

`2(-1)^2+3(-1)=2-3=-1 <0`

Thus `2x^2+3x<0` on `(-3/2,0)`

`2(1)^2+3(1) = 2+3 = 5>0`

Thus `2x^2+3x >0` on `(0,oo)`

Putting it together, then,

`2x^2+3x>0 <=> x in (-oo, -3/2)uu(0,oo)`