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

Dec 22, 2016

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

#### Explanation:

The difference of squares identity can be written:

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

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

I prefer not to have to do much arithmetic involving fractions, so I would pre-multiply this equation by $8$ to avoid them and get:

$0 = 8 \left(2 {x}^{2} - 3 x + 1\right)$

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

$\textcolor{w h i t e}{0} = 16 {x}^{2} - 24 x + 9 - 1$

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

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

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

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

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

Hence:

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

$\textcolor{w h i t e}{}$
Footnote

Why $8$?

$8 = 2 \cdot {2}^{2}$

The first factor of $2$ makes the leading term into a perfect square. The additional ${2}^{2}$ factor avoids us having to divide $3$ by $2$ and end up working with $\frac{1}{2}$'s and $\frac{1}{4}$'s