Proofs

Revision as of 22:53, 29 August 2016 by Darkdragon (talk | contribs) (Quadratic Formula)

Quadratic Formula

Let $ax^2+bx+c=0$. Then \[x^2+\frac{b}{a}x+\frac{c}{a}=0\] Completing the square, we get \[\left(x+\frac{b}{2a}\right)^2 +~ \frac{b^2-4ac}{4a^2}=0 \Rightarrow x~+~\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm \sqrt{b^2-4ac}}{2a}\] Simplifying, we see \[\boxed{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\]

Pythagorean Theorem