# How do you solve using the completing the square method 3x^2-6x-1=0?

Dec 26, 2016

$x = 2.155 \text{ or } x = - 0.155$

#### Explanation:

Quadratic equations are in the form $a {x}^{2} + b x + c = 0$

$\textcolor{\lim e}{3} {x}^{2} - 6 x - 1 = 0 \text{ } \leftarrow \div 3$ to make $\textcolor{\lim e}{a = 1}$
$\textcolor{\lim e}{\downarrow}$
$\textcolor{\lim e}{1} {x}^{2} - 2 x - \frac{1}{3} = 0 \text{ } \leftarrow$ move $- \frac{1}{3}$ to the RHS

${x}^{2} - 2 x \text{ } = \frac{1}{3}$

${x}^{2} - 2 x \textcolor{b l u e}{+ \square} = + \frac{1}{3} \text{ } \leftarrow \textcolor{b l u e}{\square = {\left(\frac{b}{2}\right)}^{2}}$
$\textcolor{w h i t e}{\ldots \ldots .} \textcolor{b l u e}{\downarrow}$
$\text{ "color(blue)(((-2)/2)^2 = 1= square)" } \leftarrow$ add to both sides

${x}^{2} - 2 x \textcolor{b l u e}{+ 1} = + \frac{1}{3} \textcolor{b l u e}{+ 1} \text{ } \leftarrow$this is completing the square

[Now use the identity $\textcolor{red}{{a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}}$ ]

$\textcolor{red}{{x}^{2} - 2 x + 1} = + 1 \frac{1}{3}$
$\textcolor{w h i t e}{\ldots .} \textcolor{red}{\downarrow}$
$\textcolor{red}{{\left(x - 1\right)}^{2}} = 1 \frac{1}{3} = \frac{4}{3} \text{ }$find square root both sides

$x - 1 = \pm \sqrt{\frac{4}{3}} \text{ } \leftarrow$ solve for $x$ twice

$x = + \sqrt{\frac{4}{3}} + 1 \text{ and } x = - \sqrt{\frac{4}{3}} + 1$

$x = \frac{2}{\sqrt{3}} + 1 \text{ and } x = \frac{- 2}{\sqrt{3}} + 1$

$x = 2.155 \text{ or } x = - 0.155$