# How do you solve 2x^2-6x+3>=0 using a sign chart?

Jun 28, 2017

The solution is $x \in \left(- \infty , \frac{3 - \sqrt{3}}{2}\right] \cup \left[\frac{3 + \sqrt{3}}{2} , + \infty\right)$

#### Explanation:

We need the roots of the equation

$2 {x}^{2} - 6 x + 3 = 0$

The discriminant is

$\Delta = {b}^{2} - 4 a c = {6}^{2} - 4 \cdot 2 \cdot 3 = 36 - 24 = 12$

As, $\Delta > 0$, there are 2 real roots

${x}_{1} = \frac{6 - \sqrt{12}}{2 \cdot 2} = \frac{6 - 2 \sqrt{3}}{4} = \frac{3 - \sqrt{3}}{2}$

${x}_{2} = \frac{6 + \sqrt{12}}{2 \cdot 2} = \frac{6 + \sqrt{3}}{4} = \frac{3 + \sqrt{3}}{2}$

Let our inequality be

$f \left(x\right) = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

We can build the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a 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}$${x}_{2}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

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

$\textcolor{w h i t e}{a a a a}$$x - {x}_{2}$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a 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 a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left(- \infty , \frac{3 - \sqrt{3}}{2}\right] \cup \left[\frac{3 + \sqrt{3}}{2} , + \infty\right)$
graph{2x^2-6x+3 [-4.93, 4.934, -2.465, 2.465]}