Completing the square

For all quadratic expressions, that is, of the form $ax^2 + bx + c$, which we'll call canonical forms, there exists an identical expression of the form $a(x + p)^2 + q$, which we'll call completed square forms. For example, $3x^2 + 12x + 27 \equiv 3(x + 2)^2 + 15$. Among other reasons, completed square forms are convient for finding the root of the quadratic's graph because it is easier to manipulate algebraically. Below is the method for converting a quadratic expression in canonical form into an identical quadratic in completed square form.

The trick to completing the square is to see what would happen if, given $x^2 + bx + c$, you assume that the completed square form is $(x + \frac{b}{2})^2$ and correct it. If we have an $a$ in front of $x^2$, we can factor it out of the expression so that we have $a(x^2 + \frac{b}{a}x + \frac{c}{a})$ and then complete the square of what's inside the brackets. Let's use $3x^2 + 12x + 27$ as an example:

$3x^2 + 12x + 27 \equiv 3(x^2 + 4x + 9)$

Focus on $x^2 + 4x + 9$

Assume that $x^2 + 4x + 9 \equiv (x + 2)^2$

Expanding the brackets:
$(x + 2)^2 \equiv x^2 + 2 \times 2 \times x + 2^2 \equiv x^2 + 4x + 4$

We can see that dividing $b$ (that is, 4) by 2 is useful because that term gets doubled after expanding the brackets. We can further see that instead of 9, the last term is 4. So we need to correct for this. Fortunately, we can introduce the term $q$ to our completed square form, which can be used to correct it. We correct it by adding 5 (that is, $9 - 4$) after the squared brackets, like this:

$(x + 2)^2 + 5$

Now, when we expand the brackets, we get:

$(x + 2)^2 + 5 \equiv (x^2 + 2 \times 2 \times x + 2^2) + 5 \equiv x^2 + 4x + 9$

which is exactly what we want. Therefore,

$3x^2 + 12x + 27 \equiv 3(x^2 + 4x + 9) \equiv 3((x + 2)^2 + 5) \equiv 3(x + 2)^2 + 15$

Let's try this with the general case:

$ax^2 + bx + c \equiv a(x^2 + \frac{b}{a}x + \frac{c}{a})$

Focus on $x^2 + \frac{b}{a}x + \frac{c}{a}$

Assume that $x^2 + \frac{b}{a}x + \frac{c}{a} \equiv (x + \frac{b}{2a})^2$

$(x + \frac{b}{2a})^2 \equiv x^2 + 2 \times \frac{b}{2a} \times x + (\frac{b}{2a})^2 \equiv x^2 + \frac{b}{a}x + (\frac{b}{2a})^2$

Add $\frac{c}{a} - (\frac{b}{2a})^2$ to the completed square expression.

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

Therefore,

$$ \begin{align*} ax^2 + bx + c &\equiv a(x^2 + \frac{b}{a}x + \frac{c}{a}) \\ &\equiv a((x + \frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2) \\ &\equiv a(x + \frac{b}{2a})^2 + a\frac{c}{a} - a(\frac{b}{2a})^2 \\ &\equiv a(x + \frac{b}{2a})^2 + (c - \frac{b^2}{4a}) \end{align*} $$

You can use the last line to convert the expression directly.