Difference between revisions of "Proofs"
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==Pythagorean Theorem== | ==Pythagorean Theorem== | ||
| − | http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/image2.gif | + | [img]http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/image2.gif[/img] |
Since area of green square is <math>a^2</math> | Since area of green square is <math>a^2</math> | ||
Revision as of 13:15, 15 May 2017
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
[img]http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/image2.gif[/img]
Since area of green square is 
Since are of blue square is 
Since red square is 
We have the following relationship
Based on this, we get we get 
(here we get the Pythagorean Theorem)
