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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 , see
Section A.2 .