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

 

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

Guest Sep 16, 2017
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 #1
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Go online to this link and see how to prove the "Quadratic Formula":

https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/v/proof-of-quadratic-formula

Guest Sep 16, 2017
 #2
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 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.
\(ax^2+bx+c=0\)  
   

 

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

TheXSquaredFactor  Sep 16, 2017

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