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

##### 3 Answers
Jul 17, 2018

The solution is $x \in \left(- \infty , 1\right] \cup \left[\frac{4}{3} , + \infty\right)$

#### Explanation:

The inequality is

$\left(x - 1\right) \left(3 x - 4\right) \ge 0$

Let $f \left(x\right) = \left(x - 1\right) \left(3 x - 4\right)$

Let's build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a}$$1$$\textcolor{w h i t e}{a a a a a a a a}$$\frac{4}{3}$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x - 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$3 x - 4$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$color(white)(aaaaa)-$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a}$$+$

Therefore,

$f \left(x\right) \ge 0$,when $x \in \left(- \infty , 1\right] \cup \left[\frac{4}{3} , + \infty\right)$

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

$x \setminus \in \left(- \setminus \infty , 1\right] \setminus \cup \left[\frac{4}{3} , \setminus \infty\right)$

#### Explanation:

Given inequality

$\left(x - 1\right) \left(3 x - 4\right) \setminus \ge 0$

setting $x - 1 = 0 \setminus \implies x = 1$

setting $3 x - 4 = 0 \setminus \implies x = \frac{4}{3}$

Specify the points $x = 1$ & $x = \frac{4}{3}$ on the number line & divide it into positive & negative regions alternatively from right most value which gives us

$x \setminus \in \left(- \setminus \infty , 1\right] \setminus \cup \left[\frac{4}{3} , \setminus \infty\right)$

Jul 17, 2018

$x \in \left(- \infty , 1\right] \cup \left[\frac{4}{3} , \infty\right)$

#### Explanation:

$\text{find the zeros of left side by equating to zero}$

$\left(x - 1\right) \left(3 x - 4\right) = 0$

$x - 1 = 0 \Rightarrow x = 1$

$3 x - 4 = 0 \Rightarrow x = \frac{4}{3}$

$\left(x - 1\right) \left(3 x - 4\right) = 3 {x}^{2} - 7 x + 4 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{Since "a>0" then minimum turning point } \bigcup$

$\left(x - 1\right) \left(3 x - 4\right) \ge 0 \text{ then}$

$x \le 1 \text{ or } x \ge \frac{4}{3}$

$x \in \left(- \infty , 1\right] \cup \left[\frac{4}{3} , \infty\right)$
graph{3x^2-7x+4 [-5, 5, -2.5, 2.5]}