# How do you solve the inequality x^2+x-20 >= 0 ?

## The answer key says $- 5 \le x \le 4$

Sep 1, 2017

It seems the answer key was for ${x}^{2} + x - 20 \le 0$.

The given problem has solution:

$x \le - 5$ or $x \ge 4$

#### Explanation:

Given:

${x}^{2} + x - 20 \ge 0$

This factors as:

$\left(x + 5\right) \left(x - 4\right) \ge 0$

This will hold if any of the following:

• (x+5 > 0" " and $\text{ "x-4 > 0)" "<=>" } x > 4$

• (x+5 < 0" " and $\text{ "x-4 < 0)" "<=>" } x < - 5$

• $x + 5 = 0 \text{ "<=>" } x = - 5$

• $x - 4 = 0 \text{ "<=>" } x = 4$

Hence we find the solution set is:

$x \in \left(- \infty , - 5\right] \cup \left[4 , \infty\right)$

That is:

$x \le - 5 \text{ }$ or $\text{ } x \ge 4$

It seems that the answer key was for a different problem, namely:

${x}^{2} + x - 20 \le 0$