# Why is completing the square useful?

##### 1 Answer
Mar 8, 2018

To simplify quadratic expressions so that they become solvable with square roots.

#### Explanation:

Completing the square is an example of a Tschirnhaus transformation - the use of a substitution (albeit implicitly) in order to reduce a polynomial equation to simpler form.

So given:

$a {x}^{2} + b x + c = 0 \text{ }$ with $a \ne 0$

we could write:

$0 = 4 a \left(a {x}^{2} + b x + c\right)$

$\textcolor{w h i t e}{0} = 4 {a}^{2} {x}^{2} + 4 a b x + 4 a c$

$\textcolor{w h i t e}{0} = {\left(2 a x\right)}^{2} + 2 \left(2 a x\right) b + {b}^{2} - \left({b}^{2} - 4 a c\right)$

$\textcolor{w h i t e}{0} = {\left(2 a x + b\right)}^{2} - {\left(\sqrt{{b}^{2} - 4 a c}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(2 a x + b\right) - \sqrt{{b}^{2} - 4 a c}\right) \left(\left(2 a x + b\right) + \sqrt{{b}^{2} - 4 a c}\right)$

$\textcolor{w h i t e}{0} = \left(2 a x + b - \sqrt{{b}^{2} - 4 a c}\right) \left(2 a x + b + \sqrt{{b}^{2} - 4 a c}\right)$

Hence:

$2 a x = - b \pm \sqrt{{b}^{2} - 4 a c}$

So:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

So having started with a quadratic equation in the form:

$a {x}^{2} + b x + c = 0$

we got it into a form ${t}^{2} - {k}^{2} = 0$ with $t = \left(2 a x + b\right)$ and $k = \sqrt{{b}^{2} - 4 a c}$, eliminating the linear term leaving only squared terms.

So long as we are happy calculating square roots, we can now solve any quadratic equation.

Completing the square is also useful for getting the equation of a circle, ellipse or other conic section into standard form.

For example, given:

${x}^{2} + {y}^{2} - 4 x + 6 y - 12 = 0$

completing the square we find:

${\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} = {5}^{2}$

allowing us to identify this equation as that of a circle with centre $\left(2 , - 3\right)$ and radius $5$.