Proofs

Revision as of 18:56, 10 September 2016 by Burunduchok (talk | contribs) (Pythagorean Theorem)

Quadratic Formula

Let $ax^2+bx+c=0$. Then \[x^2+\frac{b}{a}x+\frac{c}{a}=0\] Completing the square, we get \[\left(x+\frac{b}{2a}\right)^2 +~ \frac{b^2-4ac}{4a^2}=0 \Rightarrow x~+~\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm \sqrt{b^2-4ac}}{2a}\] Simplifying, we see \[\boxed{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\]

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

Based on this, we get we get $\boxed {a^2+b^2=c^2}$ (here we get the Pythagorean Theorem)

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