How do you solve #(x-1)(3x-4)>=0#?

3 Answers
Jul 17, 2018

Answer:

The solution is #x in (-oo, 1] uu[4/3,+oo)#

Explanation:

The inequality is

#(x-1)(3x-4)>=0#

Let #f(x)=(x-1)(3x-4)#

Let's build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##1##color(white)(aaaaaaaa)##4/3##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x-1##color(white)(aaaaa)##-##color(white)(aaaa)##0##color(white)(aaaa)##+##color(white)(aaaaaa)##+#

#color(white)(aaaa)##3x-4##color(white)(aaaa)##-##color(white)(aaaa)####color(white)(aaaaa)##-##color(white)(aa)##0##color(white)(aaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaa)##0##color(white)(aaaa)##-##color(white)(aa)##0##color(white)(aaa)##+#

Therefore,

#f(x)>=0#,when #x in (-oo, 1] uu[4/3,+oo)#

graph{(x-1)(3x-4) [-1.453, 3.413, -0.497, 1.936]}

Answer:

#x\in(-\infty, 1]\cup[4/3, \infty)#

Explanation:

Given inequality

#(x-1)(3x-4)\ge0#

setting #x-1=0\implies x=1#

setting #3x-4=0\implies x=4/3#

Specify the points #x=1# & #x=4/3# on the number line & divide it into positive & negative regions alternatively from right most value which gives us

#x\in(-\infty, 1]\cup[4/3, \infty)#

Jul 17, 2018

Answer:

#x in(-oo,1]uu[4/3,oo)#

Explanation:

#"find the zeros of left side by equating to zero"#

#(x-1)(3x-4)=0#

#x-1=0rArrx=1#

#3x-4=0rArrx=4/3#

#(x-1)(3x-4)=3x^2-7x+4larrcolor(blue)"in standard form"#

#"Since "a>0" then minimum turning point "uuu#

#(x-1)(3x-4)>=0" then"#

#x<=1" or "x>=4/3#

#x in(-oo,1]uu[4/3,oo)#
graph{3x^2-7x+4 [-5, 5, -2.5, 2.5]}