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

1 Answer
Apr 9, 2017

Answer:

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

Explanation:

#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]