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# Solve for x

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+1860

$$ax^2+bx+c=0$$; Solve for x.

gibsonj338  Nov 20, 2016
#1
+93691
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$$ax^2+bx+c=0$$

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

that is it!

Melody  Nov 20, 2016
#2
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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
+13087
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Here is a webpage with the derivation of the Quadratic Formula:

scroll down a little bit to see the derivation

ElectricPavlov  Nov 20, 2016
#4
+93691
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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