We can use extraction of roots to solve equations of the form
\begin{equation*}
a(x + p)^2 + q = 0
\end{equation*}
where the left side of the equation includes the square of a binomial, or a perfect square. It turns out that we can write any quadratic equation in this form.
In each case, the square of the binomial is a quadratic trinomial,
\begin{equation*}
(x + p)^2= x^2 + 2px + p^2
\end{equation*}
The coefficient of the linear term, \(2p\text{,}\) is twice the constant in the binomial, and the constant term of the trinomial, \(p^2\text{,}\) is its square. \(~\alert{\text{[TK]}}\)
We would like to reverse the process and write a quadratic expression as the square of a binomial. For example, what constant term can we add to
\begin{equation*}
x^2 - 16x
\end{equation*}
to produce a perfect square trinomial? Compare the expression to the formula above:
\begin{align*}
x^2 + \alert{2p}x + p^2 \amp = (x + p)^2 \\
x^2 - \alert{16}x + ~\text{?}~ \amp = (x + ~\text{?}~)^2
\end{align*}
We see that \(~2p = -16~\text{,}\) so
\begin{equation*}
p = \frac{1}{2}(-16) = \alert{-8}~~~\text{and}~~~ p^2 = (-8)^2 = \alert{64}
\end{equation*}
We substitute these values for \(p^2\) and \(p\) into the equation to find
\begin{equation*}
x^2 - 16x + \alert{64} = (x \alert{-8})^2
\end{equation*}
You can check that in the resulting trinomial, the constant term is equal to the square of one-half the coefficient of \(x\text{.}\) In other words, we can find the constant term by taking one-half the coefficient of \(x\) and then squaring the result. Adding a constant term obtained in this way is called completing the square.