# How do you solve using the completing the square method 2x^2 - x - 5 = 0?

Mar 10, 2016

$\text{ "2(x-1/4)^2-41/8 => "vertex} \to \left(x , y\right) \to \left(\frac{1}{4} , - \frac{41}{8}\right)$

x-axis and y-axis Intercepts can be found in the normal way

#### Explanation:

For a more in depth approach have a look at my solution to
http://socratic.org/s/asD9k2Ch. Diferent values but the method is sound.

Write as:$\text{ } 2 \left({x}^{2} - \textcolor{red}{\frac{1}{2}} x\right) - 5 = 0$

For the $\textcolor{red}{- \frac{1}{2}} \text{ in } - \frac{1}{2} x$ apply: $\left(- \frac{1}{2}\right) \times \left(\textcolor{red}{- \frac{1}{2}}\right) = + \frac{1}{4}$

So the left hand side becomes:

$\text{ } 2 \left({x}^{2} - \frac{1}{4} \textcolor{g r e e n}{x}\right) - 5$

Remove the $\textcolor{g r e e n}{x}$

$\text{ } 2 \left({x}^{\textcolor{m a \ge n t a}{2}} - \frac{1}{4}\right) - 5$

Move the index (power) $\textcolor{m a \ge n t a}{2}$ to outside the brackets

$\text{ } 2 {\left(x \textcolor{red}{- \frac{1}{4}}\right)}^{\textcolor{m a \ge n t a}{2}} - 5$

Square the constant color(red)((-1/4)^2=+1/16 and subtract twice its value $\textcolor{red}{2 \times {\left(= \frac{1}{4}\right)}^{2} = + \frac{1}{8}}$

$\text{ } 2 {\left(x - \frac{1}{4}\right)}^{2} - 5 - \frac{1}{8}$

$\text{ } 2 {\left(x - \frac{1}{4}\right)}^{2} - \frac{41}{8}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~
${x}_{\text{vertex}} = \left(- 1\right) \times \left(- \frac{1}{4}\right) = + \frac{1}{4}$

${y}_{\text{vertex}} = - \frac{41}{8} = - 5 \frac{1}{8}$