Question

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

Answer 1

`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`