The quadratic formula: $$f(x)=ax^2+bx+c=0$$

$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

I remember a tutor once showing me a method for deriving the quadratic formula using calculus somehow. This was around 20 years ago and I can't even remember the tutor's name. I'd really like to learn this method. Just to clarify, I do know how to derive it using the "Completing the square" method.

I was linked to the solution here: https://www.google.com/amp/s/threesixty360.wordpress.com/2008/10/19/using-calculus-to-generate-the-quadratic-formula/amp/

But I am stuck at one step.

Start with: $$f(x)=ax^2+bx+c$$

We want: $$f(x)=0$$

The first derivative gives: $$f'(x)=2ax+b$$

Which leads to this: $$f(x)=c+\int_0^x (2at+b)dt$$

I can't see why the $t's$ were introduced here.

If anyone has any other methods I'd really like to see them also.

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    $\begingroup$ Here is a site detailing just that google.com/amp/s/threesixty360.wordpress.com/2008/10/19/… $\endgroup$ – Triatticus May 12 '17 at 3:11
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    $\begingroup$ @Triatticus it again seems like a completion of squares to me. The only role of calculus being to bring that completion of squares. I was hoping to see some sort of argument which uses zero-crossing of a function, or some argument based on slope, or something like that. Correct me if I am wrong. $\endgroup$ – dineshdileep May 12 '17 at 3:39
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    $\begingroup$ See my answer on this question. You have to scroll down a bit $\endgroup$ – MathematicsStudent1122 May 12 '17 at 3:42
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    $\begingroup$ It is definitely very much like completing the square especially in the step involving the substitution, it is however a very new take on it $\endgroup$ – Triatticus May 12 '17 at 3:46

If f is a polynomial of degree $n$ then for all $x, y$ we have $$f(x)=\sum_{j=0}^n(x-y)^jf^{(j)}(y)/j!$$ where $f^{(0)}=f$ and $f^{(j)}$ is the $j$th derivative of $f$ when $j>0.$... And with the usual convention that $0^0=1 $ (i.e. the term $(x-y)^j$ for $j=0$, when $x=y$).

When $f(x)=Ax^2 +Bx+C$ with $A \ne 0,$ then $f'(x)=2Ax+B$ is equal to $0$ when $x=x_0=-B/2A.$ For all $x$ we have $$f(x)=f(x_0)+(x-x_0)f'(x_0)+(x-x_0)^2f''(x_0)/2!.$$ But $f'(x_0)=0$ and $f''(x_0)=2A,$ so for all $x$ we have $$f(x)=f(x_0)+(x-x_0)^2\cdot A.$$ This "completes the square" for us.


Let $\alpha,\beta$ be the two roots. Then we have by comparing coefficients$$\alpha+\beta=\frac{-b}{a}$$ and also you get $$\alpha.\beta=\frac{c}{a}$$. From here you immediately get $(\alpha-\beta)^2=(\frac{b}{a})^2-4.\frac{c}{a}$ and then you solve for $\alpha, \beta$

Note: $(\alpha+\beta)^2=(\alpha-\beta)^2+4.\alpha.\beta$

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    $\begingroup$ This is NOT what the OP asked for. He clarified this part. $\endgroup$ – user2902293 May 12 '17 at 3:17
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    $\begingroup$ He asked at last if anyone has other methods. You should pay more attention $\endgroup$ – user379195 May 12 '17 at 3:19
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    $\begingroup$ Oh right. My bad. $\endgroup$ – user2902293 May 12 '17 at 3:20
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    $\begingroup$ Actually I'm not sure what you did in the last step. Why did we immediately get that final equation? $\endgroup$ – Kantura May 12 '17 at 3:30
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    $\begingroup$ @SoumikGhosh The OP asked for a calculus proof.in particular, and specifically not a complete-the-square one. Your answer is a mix of Vieta's formulas and some less than clear complete-the-square argument next, but I fail to see how it addresses the calculus request in the question. $\endgroup$ – dxiv May 12 '17 at 5:52

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