# How do you complete the square for 2x^2 + 8x?

Jul 15, 2015

Extract the coefficient of ${x}^{2}$ as a factor then add (and ultimately subtract) the square of $\frac{1}{2}$ of the coefficient of $x$

#### Explanation:

Given $2 {x}^{2} + 8 x$

Extract the coefficient of ${x}^{2}$ as a factor
$\textcolor{w h i t e}{\text{XXXX}}$$= \left(2\right) \left({x}^{2} + 4 x\right)$

Half of the coefficient of $x$ is $\frac{1}{2} \times 4 = 2$
So the square of half the coefficient of $x$ is ${2}^{2} = 4$

Add (and subtract) the square of half the coefficient of $x$
$\textcolor{w h i t e}{\text{XXXX}}$$= \left(2\right) \left({x}^{2} + 4 x + {2}^{2} - 4\right)$

Which could be written as
$\textcolor{w h i t e}{\text{XXXX}}$$= \left(2\right) \left({\left(x + 2\right)}^{2} - 4\right)$

In case you were wondering "why the square of half the coefficient of $x$?)
$\textcolor{w h i t e}{\text{XXXX}}$We are trying for a squared binomial of the form ${\left(x + a\right)}^{2}$
$\textcolor{w h i t e}{\text{XXXX}}$Since
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$
$\textcolor{w h i t e}{\text{XXXX}}$Given the first 2 terms in the form:
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + d$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$($d = 2 a$)
$\textcolor{w h i t e}{\text{XXXX}}$We need to add ${\left(\frac{d}{2}\right)}^{2}$ to get a "squared form"