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


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.


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.



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

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 $x$.

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


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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