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


Is there a method to prove this? How do I know that this isn't simply a magical formula? Or is it magical?

 Sep 16, 2017

Go online to this link and see how to prove the "Quadratic Formula":


 Sep 16, 2017

 I'll show my method of proving. Of course, the standard form of a quadratic equation is in the following form of \(ax^2+bx+c=0\). If I start with the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), then all I should have to do is utilize some algebraic manipulation to arrive to the standard form. Let's do that!

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) First, let's eliminate the fraction on the right hand side by multiplying by 2a on both sides.
\(2ax=-b\pm\sqrt{b^2-4ac}\) Add b to both sides of the equation.
\(2ax+b=\pm\sqrt{b^2-4ac}\) Square both sides. Normally, you would have to consider both cases, but squaring always results in the positive answer--no matter if the original number is positive or negative. Therefore, both will result to the same thing.
\((2ax+b)^2=b^2-4ac\) Expand the left hand side of the equation by knowing that \((x+y)^2=x^2+2xy+y^2\).
\((2ax)^2+2(2ax)(b)+b^2=b^2-4ac\) Subtract b^2 from both sides.
\((2ax)^2+2(2ax)(b)=-4ac\) Simplify the left hand side by dealing with the multiplication and the exponent.
\(4a^2x^2+4axb=-4ac\) Divide by the GCF of both sides, which is 4a. 
\(ax^2+bx=-c\) And finally, subtract c on both sides.


I have now proven that the quadratic formula works for all numbers for A,B, and C. 

 Sep 16, 2017

15 Online Users


New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.