# How do you solve and write the following in interval notation: 3a - 6a^2 >0?

Apr 9, 2017

Solution : $0 < a < \frac{1}{2}$. In interval notation: $\left(0 , \frac{1}{2}\right)$.

#### Explanation:

$3 a - 6 {a}^{2} > 0 \mathmr{and} 3 a \left(1 - 2 a\right) > 0 \mathmr{and} a \left(1 - 2 a\right) > 0$

Critical points are $a = 0$ and $a = \frac{1}{2}$

When $a < 0 , a \left(1 - 2 a\right) < 0$ (1)
When $0 < a < \frac{1}{2} , a \left(1 - 2 a\right) > 0$ (2)
When $a > \frac{1}{2} , a \left(1 - 2 a\right) < 0$ (3)

Solution : $0 < a < \frac{1}{2}$. In interval notation: $\left(0 , \frac{1}{2}\right)$. In graph also $3 a - 6 {a}^{2} > 0$ for $0 < a < \frac{1}{2}$ graph{-6x^2+3x [-10, 10, -5, 5]} [Ans]