# How do you solve the inequality x^2+ 3x>10?

Jun 18, 2015

This inequality is equivalent to $\left(x + 5\right) \left(x - 2\right) > 0$ which is satisfied for $x \in \left(- \infty , - 5\right) \cup \left(2 , \infty\right)$

#### Explanation:

First subtract $10$ from both sides to get ${x}^{2} + 3 x - 10 > 0$

$f \left(x\right) = {x}^{2} + 3 x - 10$ is a well behaved, continuous function, so the truth of the inequality will only change at the points where $f \left(x\right) = 0$.

$f \left(x\right) = {x}^{2} + 3 x - 10 = \left(x + 5\right) \left(x - 2\right)$

so $f \left(- 5\right) = f \left(2\right) = 0$

Since we are looking for $f \left(x\right) > 0$, there are only three ranges we need to look at:

$\left(- \infty , - 5\right)$ : $\left(x + 5\right) < 0$ and $\left(x - 2\right) < 0$, so $f \left(x\right) > 0$

$\left(- 5 , 2\right)$ : $\left(x + 5\right) > 0$, $\left(x - 2\right) < 0$, so $f \left(x\right) < 0$

$\left(2 , \infty\right)$ : $\left(x + 5\right) > 0$ and $\left(x - 2\right) > 0$, so $f \left(x\right) > 0$

Putting these cases together:

$f \left(x\right) > 0$ for $x \in \left(- \infty , - 5\right) \cup \left(2 , \infty\right)$