Difference between revisions of "Proofs"
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http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/image2.gif | http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/image2.gif | ||
| − | Since area of green square is a^2 | + | Since area of green square is <math>a^2</math> |
| − | Since are of blue square is b^2 | + | Since are of blue square is <math>b^2</math> |
| − | Since red square is c^2 | + | Since red square is <math>c^2</math> |
We have the following relationship | We have the following relationship | ||
| − | a^2+b^2=c^2 (here we get the | + | Based on this, we get we get <math>\boxed {a^2+b^2=c^2}</math> (here we get the Pythagorean Theorem) |
Latest revision as of 13:16, 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
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
Based on this, we get we get 
(here we get the Pythagorean Theorem)
