Difference between revisions of "Proofs"
Premchandj (talk | contribs) (→Pythagorean Theorem) |
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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) | |
Revision as of 17:55, 10 September 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 
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)
