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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#

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