Completing the Square on the Standard Form of Quadratic

04 Aug, 2024

I wanted to see what LaTeX would look like here so this is a brief progression I put together for going from the standard form of a quadratic to the vertex form using completing the square. There's more background and a geometric perspective I think is helpful here, but I wanted to try it out.

Given: $f (x) = a x^{2} + b x + c$ .
First move the $c$ constant to the other side, since we want to focus on the variables to "complete the square:"

f (x) - c = a x^{2} + b x

Factor out an $a :$

f (x) - c = a (x^{2} + \frac{b}{a} x)

Take ${((\frac{1}{2}) (\frac{b}{a}))}^{2}$ or $(\frac{b}{2 a})^{2} = \frac{b^{2}}{4 a^{2}}$ and add to both sides (adding by zero):¹

f (x) - c + a (\frac{b^{2}}{4 a^{2}}) = a (x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}})

Then factor the $x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}}$ term and simplify the $- c + a (\frac{b^{2}}{4 a^{2}})$ term:²

f (x) - k = a {(x + \frac{b}{2 a})}^{2}

Rewrite:

$f (x) = a {(x - (- \frac{b}{2 a}))}^{2} + k$
$f (x) = a (x - h)^{2} + k$ with $h = - \frac{b}{2 a}$ and $k = \frac{4 a c - b^{2}}{4 a}$

You can distribute the $a$ on the right-hand side to notice why we're multiplying by $a$ on the left-hand side.↩
we can let $k = - c + a (\frac{b^{2}}{4 a^{2}})$ since it is some constant. Also $k = \frac{4 a c - b^{2}}{4 a}$ when you add the fraction. Notice that the constant term inside the parenthesis for the right-hand side of the equation is just $\frac{b}{2 a}$ .↩

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