soajagbe

Deriving the quadratic formula.

Because I want it to remain reachable.

WorkMathBraver's Proofs

January 20, 2026

Question

Given the general quadratic equation:

ax2+bx+c=0a x^2+b x+c=0

Derive the quadratic formula:

x=−b±b2−4ac2ax=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}

Solution

∴ax2+bx+c=0x2+bax+ca=0\therefore \begin{aligned}& a x^2+b x+c=0 \\& x^2+\frac{b}{a} x+\frac{c}{a}=0\end{aligned}

If complete squares are of the form:

x2+2ax+a2=(x+a)2x^2+2 a x+a^2=(x+a)^2

Then, constant a2=[2a2]2a^2=\left[\frac{2 a}{2}\right]^2

And complete squares become:

x2+2ax+[2a2]2=(x+a)2x^2+2 a x+\left[\frac{2 a}{2}\right]^2=(x+a)^2

For the second equation above, this gives:

x2+bax+[ba×12]2+ca=[ba×12]2x^2+\frac{b}{a} x+\left[\frac{b}{ a}\times \frac1 2\right]^2+\frac{c}{a}=\left[\frac{b}{ a}\times \frac1 2\right]^2

x2+bax+b24a2+ca=b24a2x^2+\frac{b}{a} x+\frac{b^2}{4 a^2} + \frac{c}{a} =\frac{b^2}{4 a^2}

x2+bax+b24a2=b24a2−cax^2+\frac{b}{a} x+\frac{b^2}{4 a^2} =\frac{b^2}{4 a^2} - \frac{c}{a}

Now the left hand side is a perfect square. [x+b2a]2=b24a2−ca\left[x+\frac{b}{2a}\right]^2 = \frac{b^2}{4a^2}-\frac{c}{a}

x+b2a=±b24a2−cax+\frac{b}{2a} = \pm \sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}

x+b2a=±b24a2−4ac4a2x+\frac{b}{2a} = \pm \sqrt{\frac{b^2}{4a^2}-\frac{4ac}{4a^2}}

x+b2a=±b2−4ac4a2x+\frac{b}{2a} = \pm \sqrt{\frac{b^2-4ac}{4a^2}}

x+b2a=±b2−4ac2ax+\frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2a}

x=−b2a±b2−4ac2ax = \frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Voila.

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