# How do you solve  14-5x-x^2<0?

Mar 29, 2017

$x < - 7 \cup x > 2$

#### Explanation:

Just to make things easier, we multiply both sides by $- 1$ and flip the inequality sign: ${x}^{2} + 5 x - 14 > 0$. This way the coefficient before ${x}^{2}$ is positive. (Note: this is not a necessary step.)

We factorize the expression: ${x}^{2} + 5 x - 14 = \left(x + 7\right) \left(x - 2\right) > 0$ to find the roots: $x = - 7$ or $2$.

These are the only two possible points where the expression can change signs. Thus, there are three regions that have the same sign:

• $x < - 7$
• $- 7 < x < 2$
• $x > 2$

Since there are no repeated roots, the signs of the region alternate. We substitute an arbitrary number in the second region to find the sign: $x = 0$. Then ${x}^{2} + 5 x - 14 = {0}^{2} + 5 \cdot 0 - 14 = - 14 < 0$. Thus, the second region is negative.

Therefore, the first and third regions are positive (you can use substitution to be sure). Looking at the inequality, ${x}^{2} + 5 x - 14 > 0$, we notice that we need to find regions where the value is positive.

The solution is the first and third regions: $x < - 7 \cup x > 2$.