# How do you solve 2x^2-8x+3<0 using a sign chart?

Feb 24, 2017

The solution is x in ](4-sqrt10)/2, (4+sqrt10)/2[

#### Explanation:

We need the roots of the equation before building the sign chart

$2 {x}^{2} - 8 x + 3 = 0$

We calculate the discriminant

$\Delta = {b}^{2} - 4 a c = {\left(- 8\right)}^{2} - 4 \left(2\right) \left(3\right)$

$= 64 - 24 = 40$

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

${x}_{1} = \frac{8 - \sqrt{40}}{4} = \frac{4 - \sqrt{10}}{2}$

${x}_{2} = \frac{8 + \sqrt{40}}{4} = \frac{4 + \sqrt{10}}{2}$

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

We can, now, 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,

$f \left(x\right) < 0$ when x in ]x_1,x_2[, $\implies$

x in ](4-sqrt10)/2, (4+sqrt10)/2[