How do you solve x^2-x-6<=0 using a sign chart?

Jul 27, 2017

Solution : $- 2 \le x \le 3$ , in interval notation: $\left[- 2 , 3\right]$

Explanation:

${x}^{2} - x - 6 \le 0 \mathmr{and} \left(x - 3\right) \left(x + 2\right) \le 0$ . Critical points are

$x = - 2 \mathmr{and} x = 3$ . for $x = - 2 \mathmr{and} x = 3 , \left(x - 3\right) \left(x + 2\right) = 0$

Sign change:

When $x < - 2$ sign of $\left(x + 2\right) \left(x - 3\right)$ is  (-)*(-) = (+) ; (x+2)(x-3)>0

When $- 2 < x < 3$ sign of $\left(x + 2\right) \left(x - 3\right)$ is  (+)*(-) = (-) ; (x+2)(x-3)<0

When $x > 3$ sign of $\left(x + 2\right) \left(x - 3\right)$ is  (+)*(+) = (+) ; (x+2)(x-3) >0

Solution : $- 2 \le x \le 3$ , in interval notation $\left[- 2 , 3\right]$ [Ans]