Completing the Square

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Section A.1 Completing the Square

An important algebraic technique is known as completing the square. For example, when trying to identify the shape of the curve satisfying

\begin{equation} x^2 - 6x + y^2 = 16\tag{A.1.1} \end{equation}

we can start by noticing the terms \(x^2+y^2\text{,}\) which suggests that this implicit equation might describe a circle. The equation for a circle involves the sum of squares, so we need to find a way to convert the terms \(x^2-6x\) into a perfect square. That is, we'd like to replace these terms by something of the form \((x-h)^2\text{.}\) Expanding, we have

\begin{equation} (x-h)^2 = x^2 - 2hx + h^2\tag{A.1.2} \end{equation}

so it is clear that we should set \(h=3\text{.}\) What about the missing term \(h^2=9\text{?}\) Take it from the constant term \(16\text{!}\) Explicitly, since

\begin{equation} (x-3)^2 = x^2 - 6x + 9\tag{A.1.3} \end{equation}

we have

\begin{equation} 16 = x^2 - 6x + y^2 = \bigl((x-3)^2 - 9\bigr) + y^2\tag{A.1.4} \end{equation}

or in other words

\begin{equation} (x-3)^2 + y^2 = 25\tag{A.1.5} \end{equation}

which is the equation of a circle of radius \(5\) centered at the point \((3,0)\text{.}\)

This example demonstrates the technique known as completing the square. Given an expression of the form

\begin{equation} Q = ax^2 + bx\tag{A.1.6} \end{equation}

we'd like to eliminate the linear term. Reasoning along similar lines as above, we have

\begin{align} Q \amp= a \left(x^2 + \frac{b}{a}x \right) = a \left( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} \right)\notag\\ \amp= a \left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a}\tag{A.1.7} \end{align}

Completing the square is the technique used to prove the quadratic formula, as we can now verify. Starting from

\begin{equation} ax^2 + bx + c = 0\tag{A.1.8} \end{equation}

we immediately have

\begin{equation} a \left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c = 0\tag{A.1.9} \end{equation}

so that

\begin{equation} \left(x+\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2-4ac}{4a^2} .\tag{A.1.10} \end{equation}

Taking the square root of both sides and rearranging terms yields the quadratic formula in its standard form.