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# Proof of the Quadratic Formula

Posted Feb23-10 at 08:08 PM by jhae2.718
Updated May19-11 at 06:07 PM by jhae2.718

I was reading the Library article on quadratic equations--I'm not sure why--and noticed that a proof of the quadratic formula is not offered. Here is a proof that depresses the quadratic into an equation of the form ax2=b, where a and b are some arbitrary constants.

For the general quadratic equation ax2+bx+c=0:

Let $$x=y-\frac{b}{2a}$$
$$a \left(y-\frac{b}{2a} \right)^2 +b\left(y-\frac{b}{2a}\right)+c=0$$
Expand $$\left(y-\frac{b}{2a}\right)^2$$ and multiply by a:
$$ay^2-by+\frac{b^2}{4a}+by-\frac{b^2}{2a}+c=0$$
Eliminate the by terms
\begin{align*} ay^2+\frac{b^2}{4a}-\frac{b^2}{2a}+c &= 0 \\ ay^2 &= \frac{b^2}{2a}-\frac{b^2}{4a}-c \\ y^2 &= \frac{b^2}{2a^2}-\frac{b^2}{4a^2}-\frac{c}{a} \\ y^2 &= \frac{b^2-4ac}{4a^2} \\ y &= \frac{\pm\sqrt{b^2-4ac}}{2a} \end{align*}

Since x=y-b/(2a),
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Q.E.D.

Personally, I find this method more appealing than completing the square.
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1.  Nicely done. I have a couple of comments: 1. Quote: Quote by jhae2.718 Personally, I find this method more appealing than completing the square. I too find it appealing, having had calculus. Calculus tells us that the extremum point is at x=-b/2a, thus motivating the variable transformation you did. That being said, I am still in favor of completing the square when teaching high school algebra students. 2. I don't think derivations are necessary in the library, and in fact could bog down many entries if that were to be a requirement. But I have gone ahead and posted a link to this blog entry, in the "See Also" section of the quadradic equation library entry. Posted Feb25-10 at 07:20 PM by Redbelly98 Updated Feb25-10 at 07:38 PM by Redbelly98
2.  Quote: Quote by Redbelly98 Nicely done. Thank you. Quote: Quote by Redbelly98 2. I don't think derivations are necessary in the library, and in fact could bog down many entries if that were to be a requirement. But I have gone ahead and posted a link to this blog entry, in the "See Also" section of the quadradic equation library entry. I agree; I was just reading the comments on the page that discussed proving the general solution using completing the square, and thought I'd write this proof up (admittedly, I just wanted an excuse for writing the proof other than "nerdiness" ). Posted Feb26-10 at 07:34 AM by jhae2.718 Updated Feb26-10 at 12:46 PM by jhae2.718
3.  this proof is nice, but the method of completing the square is way easer to comprehend Posted Jan23-11 at 11:58 AM by theprofessor0
4.  An innovative method. Good one! Posted Feb10-11 at 04:39 AM by khil_phys
5.  Quote: Quote by khil_phys An innovative method. Good one! While I appreciate the sentiment, I would be remiss if I did not point out that this method is not innovation on my part. This general method of "depressing" polynomials has existed for quite some time. Posted Feb10-11 at 05:36 PM by jhae2.718
6.  Poor polynomials... they must be nearly suicidal after all these years. Posted Mar15-11 at 09:27 PM by TylerH