Difference between revisions of "Proofs"
Premchandj (talk | contribs) (→Pythagorean Theorem) |
(→Pythagorean Theorem) |
||
| (4 intermediate revisions by 3 users not shown) | |||
| Line 20: | Line 20: | ||
We have the following relationship | We have the following relationship | ||
| − | + | 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)
