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

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Answer 1 · 2017-05-20T12:23:35

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

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

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]

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