Difference between revisions of "Proofs"
Darkdragon (talk | contribs) (→Quadratic Formula) |
Darkdragon (talk | contribs) (→Pythagorean Theorem) |
||
| Line 9: | Line 9: | ||
==Pythagorean Theorem== | ==Pythagorean Theorem== | ||
| + | |||
| + | http://jwilson.coe.uga.edu/emt668/emt668.student.folders/headangela/essay1/image2.gif | ||
| + | |||
| + | Since area of green square is a^2 | ||
| + | |||
| + | Since are of blue square is b^2 | ||
| + | |||
| + | Since red square is c^2 | ||
| + | |||
| + | We have the following relationship | ||
| + | |||
| + | a^2+b^2=c^2 (here we get the pythagorean theorem) | ||
Revision as of 23:57, 29 August 2016
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 a^2
Since are of blue square is b^2
Since red square is c^2
We have the following relationship
a^2+b^2=c^2 (here we get the pythagorean theorem)
