# How do you solve 8x^2 - 6x + 1 = 0 using the quadratic formula?

Jan 8, 2017

First, we "play with multipliers of $8$ (1x8, 2x4, 4x2 8x1) and 1 to factor and then solve each term for $0$:

$8 {x}^{2} - 6 x + 1 = 0$

$\left(4 x - 1\right) \left(2 x - 1\right) = 0$

Now we solve each term of the factored quadratic for $0$:

Solution 1)

$4 x - 1 = 0$

$4 x - 1 + \textcolor{red}{1} = 0 + \textcolor{red}{1}$

$4 x - 0 = 1$

$4 x = 1$

$\frac{4 x}{\textcolor{red}{4}} = \frac{1}{\textcolor{red}{4}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} x}{\cancel{\textcolor{red}{4}}} = \frac{1}{4}$

$x = \frac{1}{4}$

Solution 2)

$2 x - 1 = 0$

$2 x - 1 + \textcolor{red}{1} = 0 + \textcolor{red}{1}$

$2 x - 0 = 1$

$2 x = 1$

$\frac{2 x}{\textcolor{red}{2}} = \frac{1}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} = \frac{1}{2}$

$x = \frac{1}{2}$