How do you solve and write the following in interval notation: #6x^2+3x<4-2x#?

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Answer 1 · 2016-05-19T17:17:03

Any x, sans #-4/3 and 1/2#.

A rearrangement gives

#6(x^2+(5/6)x-2/3)<0#.

So, #x^2+(5/6)x-2/3#

#=(x+5/12)^2-(11/12)^2#

#=((x+5/12)-11/12)((x+5/12)+11/12)#

#=(x-1/2)(x+4/3)<0#

For the product to be negative, the factors should be of opposite signs.

So, #x>1/2 and <-4/3.# Also, #x<1/2 and > -4/3#.

Thus, the answer is any x, sans #1/2 and -4/3#..