Completing the Square

ax^{2}+bx+c=0

We divide all terms by the coecficient of the term with x to the second power:

x^{2}+\dfrac{bx}{a}+\dfrac{c}{a}=0

We move the term with x to the zero power to the other side of the equation

x^{2}+\dfrac{bx}{a}= -\dfrac{c}{a}

A term is added to both sides of the equation so that we can do a complete square on the left side.

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

We can now write a completed square on the left side:

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

We need to change terms on the right side so that each term has a denominator of 4a^{2}:

(x + \dfrac{b}{2a})^{2} = -\dfrac{4ac}{4a^{2}}+\dfrac{b^{2}}{4a^{2}}

We can now take square roots of the equality, taking care to not forget that a plus/minus must be included since, for example, 49 comes from both squaring 7 and squaring -7. Also, we don’t need to do it for both sides for a=\pm b will suffice:

\sqrt{(x + \dfrac{b}{2a})^{2}} = \pm\sqrt{-\dfrac{4ac}{4a^{2}}+\dfrac{b^{2}}{4a^{2}}}

And then remove a root from the left side, :

(x + \dfrac{b}{2a}) = \pm\sqrt{-\dfrac{4ac}{4a^{2}}+\dfrac{b^{2}}{4a^{2}}}

And take a factor outside of the root on the right side:

(x + \dfrac{b}{2a}) = \pm\dfrac{1}{2a} \sqrt{-\dfrac{4ac}{1}+\dfrac{b^{2}}{1}}

Denominators may be removed from the root on the right:

(x + \dfrac{b}{2a}) = \pm\dfrac{1}{2a} \sqrt{-4ac+b^{2}}

The pieces on the right side may be made into a fraction:

(x + \dfrac{b}{2a}) = \pm\dfrac{\sqrt{-4ac+b^{2}}}{2a}

The association on the left isn’t needed:

x + \dfrac{b}{2a}= \pm\dfrac{\sqrt{-4ac+b^{2}}}{2a}

Moving a term from the left to the right isolates ‘x’, giving us an acceptable definition for ‘x’:

x = -\dfrac{b}{2a} \pm\dfrac{\sqrt{-4ac+b^{2}}}{2a}

The terms on the right combine:

x = \dfrac{-b\pm\sqrt{-4ac+b^{2}}}{2a}

Changing the order of the terms within the root is the last thing we need to do in order to match the form taught to students:

x = \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}

The Cock Agent

Addition Axiom of Equality