Completing the Square

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Theorem

Let $a, b, c, x$ be real numbers with $a \ne 0$.


Then:

$a x^2 + b x + c = \dfrac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a}$


This process is known as completing the square.


Proof

\(\ds a x^2 + b x + c\) \(=\) \(\ds \frac {4 a^2 x^2 + 4 a b x + 4 a c} {4 a}\) multiplying top and bottom by $4 a$
\(\ds \) \(=\) \(\ds \frac {4 a^2 x^2 + 4 a b x + b^2 + 4 a c - b^2} {4 a}\)
\(\ds \) \(=\) \(\ds \frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a}\)

$\blacksquare$


Also presented as

This result can also be presented in the form:

$a x^2 + b x + c = \dfrac {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } {4 a}$


Also see


Historical Note

The technique of Completing the Square was known to the ancient Babylonians as early as $1600$ BCE.


Sources