Question

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

Answer 1

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 2

`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)`

Answer 3

`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]}