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

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Answer 1 · 2017-07-27T12:06:30

Solution : $-2 <= x <= 3$ , in interval notation: $[-2,3]$

$x^2 -x- 6 <= 0 or (x-3)(x+2) <= 0$ . Critical points are

$x=-2 and x=3$ . for $x=-2 or x=3 , (x-3)(x+2) = 0$

Sign change:

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

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

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

Solution : $-2 <= x <= 3$ , in interval notation $[-2,3]$ [Ans]

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