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

Jul 28, 2015

#### Answer:

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

#### Explanation:

The first thing you need to do is get your quadratic to the form

${x}^{2} + \frac{b}{a} x = - \frac{c}{a}$

To do that, add $- 1$ to both sides of the equation and divide every thing by 2, the coefficient of ${x}^{2}$.

$2 {x}^{2} - 3 x + \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} = - 1$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} - \frac{3}{2} x = - \frac{1}{2}$

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

The idea behing using the completing the square technique is that you need to write the left side of the equation as a square of a binomial by adding a term to both sides of the equation.

This term can be found by dividing the coefficient of the $x$-term by 2, then squaring it. In your case, you have

$\left(- \frac{3}{2}\right) \cdot \frac{1}{2} = - \frac{3}{4}$, then

${\left(- \frac{3}{4}\right)}^{2} = \frac{9}{16}$

Your quadratic now becomes

${x}^{2} - \frac{3}{2} x + \frac{9}{16} = - \frac{1}{2} + \frac{9}{16}$

The left side of the equation can now be written as

${x}^{2} - \frac{3}{2} x + \frac{9}{16} = {x}^{2} - 2 \cdot \frac{3}{4} x + {\left(- \frac{3}{4}\right)}^{2} = {\left(x - \frac{3}{4}\right)}^{2}$

This means that you have

${\left(x - \frac{3}{4}\right)}^{2} = \frac{1}{16}$

Take the square root of boths sides to get

$\sqrt{{\left(x - \frac{3}{4}\right)}^{2}} = \sqrt{\frac{1}{16}}$

$x - \frac{3}{4} = \pm \frac{1}{4} \implies {x}_{1 , 2} = \frac{3}{4} \pm \frac{1}{4}$

The two solutions of the quadratic will thus be

${x}_{1} = \frac{3}{4} - \frac{1}{4} = \textcolor{g r e e n}{\frac{1}{2}}$ and ${x}_{2} = \frac{3}{4} + \frac{1}{4} = \textcolor{g r e e n}{1}$