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

Question

2237 views around the world

You can reuse this answer
Creative Commons License

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]