How do you solve #6(y^2-2)+y<0# using a sign chart?

1 Answer
May 20, 2017

Answer:

Solution : # -3/2< y < 4/3 #. In interval notation # ( -3/2,4/3) #

Explanation:

#6y^2 -12+y <0 or 6y^2+y-12 < 0 or (3y-4)(2y+3)< 0 #

Critical points are # 3y-4=0 or y= 4/3, 2y+3=0 or y=-3/2#

When # y < -3/2 # sign of #(3y-4)(2y+3) is (-)*(-) = (+)# i.e #>0#

When # -3/2< y < 4/3 # sign of #(3y-4)(2y+3) is (-)*(+) = (-)# i.e #<0#

When # y > 4/3# sign of #(3y-4)(2y+3) is (+)*(+) = (+)# i.e #>0#

So Solution : # -3/2< y < 4/3 #. In interval notation # ( -3/2,4/3) #
graph{6x^2+x-12 [-40, 40, -20, 20]}
The graph also confirms above result #[Ans]