How do you solve 4x^2-8x+3<0?

1 Answer
Aug 13, 2016

color(green)(x in (1/2,3/2)

Explanation:

4x^2-8x+3 < 0
can be factored as
(2x-1)(2x-3) < 0

{: (2x-1 < 0,color(white)("XXX"),2x-3 < 0), ("if " x < 1/2,,"if " x < 3/2) :}

(2x-1)(2x-3) will be less than zero if one but not both of the terms are less than zero.

That is (2x-1)(2x-3) < 0
color(white)("XXXXXXXXXXXXXX")if 1/2 < x < 3/2