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Proof of the Quadratic Formula
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 [tex]x=y-\frac{b}{2a}[/tex]
[tex]a \left(y-\frac{b}{2a} \right)^2 +b\left(y-\frac{b}{2a}\right)+c=0[/tex]
Expand [tex]\left(y-\frac{b}{2a}\right)^2[/tex] and multiply by a:
[tex]ay^2-by+\frac{b^2}{4a}+by-\frac{b^2}{2a}+c=0[/tex]
Eliminate the by terms
[tex]
\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*}
[/tex]
Since x=y-b/(2a),
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Q.E.D.
Personally, I find this method more appealing than completing the square.
For the general quadratic equation ax2+bx+c=0:
Let [tex]x=y-\frac{b}{2a}[/tex]
[tex]a \left(y-\frac{b}{2a} \right)^2 +b\left(y-\frac{b}{2a}\right)+c=0[/tex]
Expand [tex]\left(y-\frac{b}{2a}\right)^2[/tex] and multiply by a:
[tex]ay^2-by+\frac{b^2}{4a}+by-\frac{b^2}{2a}+c=0[/tex]
Eliminate the by terms
[tex]
\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*}
[/tex]
Since x=y-b/(2a),
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Q.E.D.
Personally, I find this method more appealing than completing the square.
Total Comments 6
Comments
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Nicely done. I have a couple of comments:
1.
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.Quote:Quote by jhae2.718Personally, I find this method more appealing than completing the square.
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 -
Thank you.Quote:Quote by Redbelly98
Nicely done.
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" ).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.Posted Feb26-10 at 07:34 AM by jhae2.718
Updated Feb26-10 at 12:46 PM by jhae2.718 -
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
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An innovative method. Good one!
Posted Feb10-11 at 04:39 AM by khil_phys
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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.Quote:Quote by khil_phys
An innovative method. Good one!
Posted Feb10-11 at 05:36 PM by jhae2.718
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Poor polynomials... they must be nearly suicidal after all these years.
Posted Mar15-11 at 09:27 PM by TylerH
