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

Question

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

Eli Assouline · Accepted Answer

Solution : # 0 < a < 1/2 #. In interval notation: #(0 , 1/2)#.

#3a -6a^2 >0 or 3a(1-2a) > 0 or a(1-2a) >0# Critical points are #a=0# and #a=1/2# When #a < 0 , a (1-2a) < 0 # (1) When # 0 < a < 1/2 , a (1-2a) > 0# (2) When # a >1/2 , a (1-2a) < 0 # (3) Solution : # 0 < a < 1/2 #. In interval notation: #(0,1/2)#. In graph also #3a-6a^2>0# for # 0 < a < 1/2# graph{-6x^2+3x [-10, 10, -5, 5]} [Ans]

Jameson Allen · Answer

undefined To solve and write the inequality $3a - 6a^2 > 0$ in interval notation, we first factor out the common term $3a$ to get $3a(1 - 2a) > 0$. Then, we find the critical points by setting each factor equal to zero: $3a = 0$ and $1 - 2a = 0$, which gives us $a = 0$ and $a = \frac{1}{2}$. These critical points divide the number line into three intervals: $(-∞, 0)$, $(0, \frac{1}{2})$, and $(\frac{1}{2}, +∞)$.

Next, we test each interval by choosing a test point from each interval and plugging it into the inequality $3a - 6a^2 > 0$. For example, in the interval $(-∞, 0)$, we can choose $a = -1$ as a test point. Plugging $a = -1$ into the inequality gives $3(-1) - 6(-1)^2 = -3 - 6(1) = -9$, which is less than 0, so this interval is not part of the solution.

In the interval $(0, \frac{1}{2})$, we can choose $a = \frac{1}{4}$ as a test point. Plugging $a = \frac{1}{4}$ into the inequality gives $3(\frac{1}{4}) - 6(\frac{1}{4})^2 = \frac{3}{4} - \frac{6}{16} = \frac{3}{4} - \frac{3}{8} = \frac{3}{8}$, which is greater than 0, so this interval is part of the solution.

Finally, in the interval $(\frac{1}{2}, +∞)$, we can choose $a = 1$ as a test point. Plugging $a = 1$ into the inequality gives $3(1) - 6(1)^2 = 3 - 6 = -3$, which is less than 0, so this interval is not part of the solution.

Therefore, the solution to $3a - 6a^2 > 0$ in interval notation is $(0, \frac{1}{2})$.

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

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