# How do you solve the equation 2x^2-3x+1=0 by completing the square?

Jan 8, 2017

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

#### Explanation:

$f \left(x\right) = 2 {x}^{2} - 3 x + 1$

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We will use this with $a = \left(4 x - 3\right)$ and $b = 1$.

First pre-multiply by $8$ to avoid the need to do arithmetic with fractions:

$0 = 8 f \left(x\right)$

$\textcolor{w h i t e}{0} = 8 \left(2 {x}^{2} - 3 x + 1\right)$

$\textcolor{w h i t e}{0} = 16 {x}^{2} - 24 x + 8$

$\textcolor{w h i t e}{0} = {\left(4 x\right)}^{2} - 2 \left(4 x\right) \left(3\right) + 9 - 1$

$\textcolor{w h i t e}{0} = {\left(4 x - 3\right)}^{2} - {1}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(4 x - 3\right) - 1\right) \left(\left(4 x - 3\right) + 1\right)$

$\textcolor{w h i t e}{0} = \left(4 x - 4\right) \left(4 x - 2\right)$

$\textcolor{w h i t e}{0} = \left(4 \left(x - 1\right)\right) \left(2 \left(2 x - 1\right)\right)$

$\textcolor{w h i t e}{0} = 8 \left(x - 1\right) \left(2 x - 1\right)$

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