Proofs
Quadratic Formula
Let 
. Then
![\[x^2+\frac{b}{a}x+\frac{c}{a}=0\]](http://latex.artofproblemsolving.com/4/7/9/4799da66026f143e156a06aef1b980f2c24eeda6.png)
Completing the square, we get
![\[\left(x+\frac{b}{2a}\right)^2 +~ \frac{b^2-4ac}{4a^2}=0 \Rightarrow x~+~\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm \sqrt{b^2-4ac}}{2a}\]](http://latex.artofproblemsolving.com/9/4/0/9406b60ac893d9e1e488a33bbc34ac34a454a103.png)
Simplifying, we see
![\[\boxed{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\]](http://latex.artofproblemsolving.com/a/5/8/a5876f53aaba5c570cdf36f1e21a401e94d8e6ab.png)
Pythagorean Theorem
http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/image2.gif
Since area of green square is 
Since are of blue square is 
Since red square is 
We have the following relationship
we get we get(here we get the pythagorean theorem)
(here we get the pythagorean theorem)
