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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?

Sep 16, 2017

#1
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Go online to this link and see how to prove the "Quadratic Formula":

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.

Sep 16, 2017