Difference between revisions of "Proofs"
Darkdragon (talk | contribs) (→Pythagorean Theorem) |
Premchandj (talk | contribs) (→Pythagorean Theorem) |
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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 | ||
we get we get <math>\boxed {a^2+b^2=c^2}</math> (here we get the pythagorean theorem) | we get we get <math>\boxed {a^2+b^2=c^2}</math> (here we get the pythagorean theorem) | ||
Revision as of 17:40, 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
we get we get(here we get the pythagorean theorem)
(here we get the pythagorean theorem)
