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「平方完成する」とは、\( a\neq 0 \) のとき，\( ax^{2}+bx+c\) を \( a(x+p)^{2}+q \)　のように式を変形することである．実際に、 \( p, q \) が\( a, b, c\) を使ってどのように表されるか確かめてみよう．

\[
\begin{eqnarray*}
ax^{2}+bx+c & = & a\bigg(x^{2}+\displaystyle \frac{b}{a}x\bigg)+c \\
& = & a\displaystyle \Bigg\{x^{2}+2\cdot\frac{b}{2a}x+\bigg(\frac{b}{2a}\bigg)^{2}-\bigg(\frac{b}{2a}\bigg)^{2}\Bigg\}+c \\
& = & a\displaystyle \Bigg\{\bigg(x+\frac{b}{2a}\bigg)^{2}-\bigg(\frac{b}{2a}\bigg)^{2}\Bigg\}+c \\
& = & a\bigg(x+\displaystyle \frac{b}{2a}\bigg)^{2}-\frac{b^{2}}{4a}+c \\
& = & a\bigg(x+\displaystyle \frac{b}{a}\bigg)^{2}-\frac{b^{2}-4ac}{4a}
\end{eqnarray*}
\]

例：
\[
\begin{eqnarray*}
x^{2}+3x+5 & = & \bigg(x^{2}+2\displaystyle \cdot\frac{3}{2}x\bigg)+5 \\
& = & \displaystyle \Bigg\{\ x^{2}\ +\ 2\cdot\frac{3}{2}x\ +\bigg(\frac{3}{2}\bigg)^{2}-\bigg(\frac{3}{2}\bigg)^{2}\Bigg\}+5 \\
& = & \displaystyle \Bigg\{ \bigg(x+\frac{3}{2}\bigg)^{2}-\frac{9}{4}\Bigg\}+5 \\
& = & \bigg(x+\displaystyle \frac{3}{2}\bigg)^{2}+\frac{11}{4}
\end{eqnarray*}
\]

例：Wolfram|Alpha
\[
\begin{eqnarray*}
5x^{2}+3x+1 & = & 5\bigg(x^{2}+\displaystyle \frac{3}{5}x\bigg)+1 \\
& = & 5\displaystyle \Bigg\{x^{2}+2\cdot\frac{3}{2\cdot 5}x+\bigg(\frac{3}{2\cdot 5}\bigg)^{2}-\bigg(\frac{3}{2\cdot 5}\bigg)^{2}\Bigg\}+1 \\
& = & 5\displaystyle \Bigg\{x^{2}+2\cdot\frac{3}{10}x+\bigg(\frac{3}{10}\bigg)^{2}-\bigg(\frac{3}{10}\bigg)^{2}\Bigg\}+1 \\
& = & 5\displaystyle \Bigg\{\bigg(x+\frac{3}{10}\bigg)^{2}-\frac{9}{100}\Bigg\}+1 \\
& = & 5\Bigg(x+\displaystyle \frac{3}{10}\Bigg)^{2}-\frac{9}{20}+1 \\
& = & 5\Bigg(x+\displaystyle \frac{3}{10}\Bigg)^{2}+\frac{11}{20}
\end{eqnarray*}
\]