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

Jul 23, 2016

$2 {x}^{2} + 3 x > 0 \iff x \in \left(- \infty , - \frac{3}{2}\right) \cup \left(0 , \infty\right)$

#### 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 $0$:

$2 {x}^{2} + 3 x = x \left(2 x + 3\right) = 0$

$\iff x = 0 \mathmr{and} 2 x + 3 = 0$

$\iff x = 0 \mathmr{and} x = - \frac{3}{2}$

Thus the intervals in question are $\left(- \infty , - \frac{3}{2}\right) , \left(- \frac{3}{2} , 0\right) , \left(0 , \infty\right)$

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

$2 {\left(- 2\right)}^{2} + 3 \left(- 2\right) = 2 \left(4\right) - 6 = 8 - 6 = 2 > 0$

Thus $2 {x}^{2} + 3 x > 0$ on $\left(- \infty , - \frac{3}{2}\right)$

$2 {\left(- 1\right)}^{2} + 3 \left(- 1\right) = 2 - 3 = - 1 < 0$

Thus $2 {x}^{2} + 3 x < 0$ on $\left(- \frac{3}{2} , 0\right)$

$2 {\left(1\right)}^{2} + 3 \left(1\right) = 2 + 3 = 5 > 0$

Thus $2 {x}^{2} + 3 x > 0$ on $\left(0 , \infty\right)$

Putting it together, then,

$2 {x}^{2} + 3 x > 0 \iff x \in \left(- \infty , - \frac{3}{2}\right) \cup \left(0 , \infty\right)$