# How do you solve 2x - 6 = -5x^2 by completing the square?

##### 1 Answer
Jul 18, 2016

$x = - 1.313$ to 3 decimal places

$x = + 0.914$ to 3 decimal places

#### Explanation:

Standard form $\to y = a {x}^{2} + b x + c$

Converting the given equation to standard form gives

$y = 0 = 5 {x}^{2} + 2 x - 6$

But $y = 0$ is a specific case

So for the general case we have:

$y = 5 {x}^{2} + 2 x - 6$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

the process of completing the square introduces an error. It ads a term to the original equation. Thus this term must be removed. This is achieved by introducing a correction.

Let $k$ be the correction value.

$\textcolor{b l u e}{\text{Step 1}}$

write as:$\text{ } y = 5 \left({x}^{2} + \frac{2}{5} x\right) - 6$

Include the correction

$y = 5 \left({x}^{2} + \frac{2}{5} x\right) - 6 + k$

At this stage $k = 0$ as we have not changed the overall values.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 2}}$

Move the power from ${x}^{2}$ to outside the brackets

$y = 5 {\left(x + \frac{2}{5} x\right)}^{2} - 6 + k \leftarrow \text{ now "k" starts to have a value}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 3}}$

Remove the $x$ from $\frac{2}{5} x$

$y = 5 {\left(x + \frac{2}{5}\right)}^{2} - 6 + k$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 4}}$

Halve the $\frac{2}{5}$

$y = 5 {\left(x + \frac{2}{10}\right)}^{2} - 6 + k$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This is the point where we find the value of $k$
The error comes from: $5 {\left(\frac{2}{10}\right)}^{2}$. This is an added term and needs to be removed.

So $5 {\left(\frac{2}{10}\right)}^{2} + k = 0$

$\implies \textcolor{red}{k = - \frac{5 \times 4}{100} \text{ "=" "-20/100" " =" } - \frac{1}{5}}$

So we now have:

$y = 5 {\left(x + \frac{2}{10}\right)}^{2} - 6 \textcolor{red}{- \frac{1}{5}}$

$\textcolor{w h i t e}{\frac{2}{2}}$

$\text{ } \textcolor{b l u e}{\underline{\overline{| \textcolor{w h i t e}{\frac{2}{2}} y = 5 {\left(x + \frac{2}{10}\right)}^{2} - \frac{31}{5} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Set $y = 0$ then we have

$+ \frac{31}{25} = {\left(x + \frac{2}{10}\right)}^{2}$

$\implies x + \frac{2}{10} = \pm \sqrt{\frac{31}{25}}$

$\implies x = - \frac{2}{10} \pm \sqrt{\frac{31}{25}}$

$\implies x = - \frac{2}{10} \pm \frac{\sqrt{31}}{5}$

$x = - 1.313$ to 3 decimal places

$x = + 0.914$ to 3 decimal places 