How do you solve #4x^2-9x+2<0#?

1 Answer
Nov 6, 2016

Answer:

The answer is # 1/4 < x < 2#

Explanation:

Let's factorise the expression #f(x=)4x^2-9x+2=(4x-1)(x-2)#
#(4x-1)(x-2)<0#
We make a sign chart to solve the problems
#color(white)(aaaaaa)##x##color(white)(aaa)##-oo##color(white)(aaa)##1/4##color(white)(aaa)##2##color(white)(aaa)##+oo#
#color(white)(aaaa)##4x-1##color(white)(aaaa)##-##color(white)(a)##0##color(white)(a)##+##color(white)(aaa)##+#
#color(white)(aaaaa)##x-2##color(white)(aaaa)##-##color(white)(aaa)##-##color(white)(a)##0##color(white)(a)##+#
#color(white)(aaaaa)##f(x)##color(white)(aaaaa)##+##color(white)(aaa)##-##color(white)(a)##0##color(white)(a)##+#

So #f(x<0)# when # 1/4 < x < 2#
graph{(4x-1)(x-2) [-7.024, 7.024, -3.51, 3.513]}