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

May 20, 2017

Solution : $- \frac{3}{2} < y < \frac{4}{3}$. In interval notation $\left(- \frac{3}{2} , \frac{4}{3}\right)$

#### Explanation:

$6 {y}^{2} - 12 + y < 0 \mathmr{and} 6 {y}^{2} + y - 12 < 0 \mathmr{and} \left(3 y - 4\right) \left(2 y + 3\right) < 0$

Critical points are $3 y - 4 = 0 \mathmr{and} y = \frac{4}{3} , 2 y + 3 = 0 \mathmr{and} y = - \frac{3}{2}$

When $y < - \frac{3}{2}$ sign of $\left(3 y - 4\right) \left(2 y + 3\right) i s \left(-\right) \cdot \left(-\right) = \left(+\right)$ i.e $> 0$

When $- \frac{3}{2} < y < \frac{4}{3}$ sign of $\left(3 y - 4\right) \left(2 y + 3\right) i s \left(-\right) \cdot \left(+\right) = \left(-\right)$ i.e $< 0$

When $y > \frac{4}{3}$ sign of $\left(3 y - 4\right) \left(2 y + 3\right) i s \left(+\right) \cdot \left(+\right) = \left(+\right)$ i.e $> 0$

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