# How do you solve using the completing the square method x^2 - 1/2 x - 3/16 = 0?

Sep 19, 2016

$x = \frac{3}{4} \text{ or } x = - \frac{1}{4}$

#### Explanation:

This involves adding the correct value to a quadratic expression to create a perfect square.

Recall: ${\left(x - 5\right)}^{2} = {x}^{2} \textcolor{\to m a \to}{- 10} x \textcolor{\to m a \to}{+ 25} \text{ } \leftarrow \textcolor{\to m a \to}{{\left(\frac{- 10}{2}\right)}^{2} = + 25}$

This relationship between $\textcolor{\to m a \to}{b \mathmr{and} c}$ will always exist.

If the value of $c$ is not the correct one, add on what you need.
However, if you have an equation add to BOTH sides.

${x}^{2} - \frac{1}{2} x - \frac{3}{16} = 0 \text{ } \leftarrow$ check: is ((-1)/2 div 2)^2 = -3/16? No

(the last term MUST be positive, because it is squared)

${x}^{2} - \frac{1}{2} x \textcolor{b l u e}{- \frac{3}{16}} = 0 \text{ } \leftarrow$ not what we want, move it to RHS

${x}^{2} \textcolor{m a \ge n t a}{- \frac{1}{2}} x \text{ +? " = color(blue)(3/16) } \leftarrow$ add on what you DO want

$\left[\textcolor{m a \ge n t a}{- \frac{1}{2}} \rightarrow \div 2 \rightarrow - \frac{1}{4} \rightarrow \text{squared} = \textcolor{red}{+ \frac{1}{16}}\right]$

x^2 - 1/2 x color(red)(+1/16) = 3/16 color(red)(+1/16) "larr add to BOTH sides

Now write the square of the binomial as (x +?)^2

${\left(x \textcolor{m a \ge n t a}{- \frac{1}{4}}\right)}^{2} = \frac{4}{16} = \frac{1}{4} \text{ } \leftarrow \textcolor{m a \ge n t a}{- \frac{1}{2} \div 2 = - \frac{1}{4}}$

Now solve the equation - isolate $x$

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

$x - \frac{1}{4} = \pm \sqrt{\frac{1}{4}} \text{ } \leftarrow$ square root both sides

$x = + \sqrt{\frac{1}{4}} + \frac{1}{4} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$

Or

$x = - \sqrt{\frac{1}{4}} + \frac{1}{4} = - \frac{1}{2} + \frac{1}{4} = - \frac{1}{4}$