Question

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

Answer 1

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]