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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
    Commented May 12, 2017 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. $\endgroup$ Commented May 12, 2017 at 3:39
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    $\begingroup$ See my answer on this question. You have to scroll down a bit $\endgroup$ Commented May 12, 2017 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
    Commented May 12, 2017 at 3:46

4 Answers 4

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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.

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It seems what's confused you is the fact that, whereas an indefinite integral uses the same variable on both sides of an equation such as $\int x^2dx=\frac13x^3+C$, you can't use $x$ as both a definite integral's limit and its integration variable, i.e. you have to write $f(x)=f(0)+\int_0^xf^\prime(t)dt$ instead of $f(x)=f(0)+\int_0^xf^\prime(x)dx$. I'm sure you can follow the rest of the proof; it's just an overly complicated way of deducing $f=0$ is equivalent to a condition of the form $w^2=k$. You may want to try substituting $v=t+b/2a$ instead to check you understand the technique.

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Let

$$ y=ax^2+bx+c \tag{1} $$

and $\tfrac{dy}{dx}=2ax+b$. Now let $g=\tfrac{dy}{dx}$ such that

$$ \begin{align*} g &=2ax+b \tag{2} \\ x &= \dfrac{g-b}{2a} \tag{3}. \end{align*} $$

Substitute $(3)$ into $(1)$ and get

$$ \begin{align*} y &=a {\left( \dfrac{g-b}{2a} \right)}^2+b {\left( \dfrac{g-b}{2a} \right)}+c \\ &= a \dfrac{g^2-gb+b^2}{4a^2} + \dfrac{bg-b^2}{2a}+c \\ &= \dfrac{g^2-gb+b^2}{4a} + \dfrac{2bg-2b^2}{4a}+ \frac{4ac}{4a} \\ y &= \dfrac{g^2-b^2 +4ac}{4a} \\ \end{align*} $$

Set $y=0$ and solve for $g$.

$$ \begin{align*} y &= \dfrac{g^2-b^2 +4ac}{4a} \\ 0 &= \dfrac{g^2-b^2 +4ac}{4a} \\ 0 &= g^2-b^2 +4ac \\ g &= \pm \sqrt{b^2-4ac} \tag{4} \\ \end{align*} $$

Now substitute $(2)$ in for $g$ and solve fo $x$.

$$ \begin{align*} g &= \pm \sqrt{b^2-4ac} \tag{4} \\ 2ax+b &= \pm \sqrt{b^2-4ac} \\ x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \end{align*} $$

Another way is: let $f(x)=ax^2+bx+c$ such that

$$ f(x)=\int_{0}^{x} \left( 2at+b \right) dt + c $$

The $t$'s are introduced here because we cannot have $x$ as the variable of integration and also as a bound. Think of them as dummy variables; all the $t$'s will become $x$'s when you evaluate the integral.

Introduce the "u" substitution where $u=2at+b \implies dt= \tfrac{1}{2a}du$.

$$ f(x)=\int_{t=0}^{t=x} \dfrac{u}{2a} du + c $$

Since we are now integrating with respect to $u$, so your new lower and upper bounds respectively become $u=b$ and $u=2ax+b$. Thus, you get

$$ \begin{align*} f(x) &=\int_{b}^{2ax+b} \dfrac{u}{2a} du + c \\ & = \dfrac{1}{2a} \left( {\dfrac{ {\left( {2ax+b} \right)}^2}{2} - \dfrac{b^2}{2}} \right) + c\\ & = \dfrac{{\left( 2ax+b \right)}^2 }{4a} - \dfrac{b^2}{4a} + \dfrac{4ac}{4a}. \\ \end{align*} $$

Set $f(x)=0$ and solve for $x$.

$$ \begin{align*} 0 & = \dfrac{{\left( 2ax+b \right)}^2 }{4a} - \dfrac{b^2}{4a} + \dfrac{4ac}{4a}. \\ 0 & = \left( 2ax+b \right)^2 - b^2 + 4ac \\ \left( 2ax+b \right)^2 & = b^2 - 4ac \\ x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ \end{align*} $$

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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. $\endgroup$ Commented May 12, 2017 at 3:17
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    $\begingroup$ He asked at last if anyone has other methods. You should pay more attention $\endgroup$
    – user379195
    Commented May 12, 2017 at 3:19
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    $\begingroup$ Oh right. My bad. $\endgroup$ Commented May 12, 2017 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
    Commented May 12, 2017 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
    Commented May 12, 2017 at 5:52

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