Completing the Square

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Theorem

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


Then

$a x^2 + b x + c = \dfrac {\left({2 a x + b}\right)^2 + 4 a c - b^2} {4 a}$


This process is known as completing the square.


Proof

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

$\blacksquare$


Also see