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\(ax^2+bx+c=0\); Solve for x.

gibsonj338  Nov 20, 2016
 #1
avatar+93691 
0

 

\(ax^2+bx+c=0\)

 

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

that is it!

Melody  Nov 20, 2016
 #2
avatar+1860 
0

Can you expand it and show how you got to \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\).  I already know that that is the answer.  I want to show why the quadradic formula works.

gibsonj338  Nov 20, 2016
 #3
avatar+13087 
+5

Here is a webpage with the derivation of the Quadratic Formula:

http://www.purplemath.com/modules/sqrquad2.htm

 

scroll down a little bit to see the derivation

ElectricPavlov  Nov 20, 2016
 #4
avatar+93691 
0

Why didn't you say so   LOL

 

\(\begin{array}{rcl}\\ax^2+bx+c&=&0\\ ax^2+bx&=&-c\\ x^2+\frac{b}{a}x&=&-\frac{c}{a}\\ x^2+\frac{b}{a}x+(\frac{b}{2a})^2&=&-\frac{c}{a}+(\frac{b}{2a})^2\\ x^2+\frac{b}{a}x+(\frac{b}{2a})^2&=&-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ (x+\frac{b}{2a})^2&=&\frac{b^2-4ac}{4a^2}\\ x+\frac{b}{2a}&=&\frac{\pm\sqrt{b^2-4ac}}{2a}\\ x&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \end{array}\)

Melody  Nov 20, 2016

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