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Context: A few years ago, when I was in senior year, I participated in the Student for a Day at my local CEGEP (in Quebec, there are just 5 years of middle and high school.A CEGEP is a school you go to to learn pre-university stuff such as calculus and biology. It's the equivalent of 12th grade in the US). Anyways, I attended a calculus lecture (I had no knowledge of calculus at that time) and I remember the prof proved the quadratic formula using calculus. Today, I'm a second-year math major with more calculus knowledge than the average CEGEP student. I tried re-finding the formula using calculus and did not manage to do it. I also checked online and the only thing I found that was not done by completing the square is a Khan Academy video that doesn't work (the thumbnail is there but the page is inexistant). Do any of you know how to solve it?

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    $\begingroup$ Can you be precise as to which formula you are talking of? $\endgroup$ – SchrodingersCat Oct 24 '15 at 17:53
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    $\begingroup$ @Aniket Is there more than one formula referred to as the quadratic formula? That can be derived by completing the square? $\endgroup$ – fleablood Oct 24 '15 at 17:59
  • $\begingroup$ Isn't the calculus based thing simply a more complicated way to complete the square? The parabola hits it's extrema at $- \frac{b}{2a}$ and is symmetric about it. So expand the square $(x + \frac{b}{2a})^2$. $\endgroup$ – stochasticboy321 Oct 24 '15 at 18:01
  • $\begingroup$ @fleablood Do you mean this .... en.wikipedia.org/wiki/Sridhara? We knew it by this name...check the link to know what name I am talking of. I think the same formula can be called in different ways by different people of different nations. Did you know about the nomenclature of Morrie's formula? If not, then please check this....en.wikipedia.org/wiki/Morrie%27s_law .I don't think many people call it by that name too if you conduct a survey.. $\endgroup$ – SchrodingersCat Oct 24 '15 at 18:03
  • $\begingroup$ Point taken. I apologize. $\endgroup$ – fleablood Oct 24 '15 at 18:30
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Consider the differential equation $$ \frac{dy}{dx}=2ax+b $$ where $a,b\in \mathbb{R}$ and with initial condition $y(0)=c$. Separating the variables yields $$ \int dy=\int (2ax+b)dx $$

Set $u=2ax+b$ in the right integral to get $$ \int (2ax+b)dx=\int \frac{1}{2a}u\,du= \frac{u^2}{4a}+C $$ Reverting the substitution, we get $y=\frac{(2ax+b)^2}{4a}+C$. Now we use the initial condition to find $C=c-\frac{b^2}{4a}$. We thus can write $$ y=\frac{(2ax+b)^2}{4a}+c-\frac{b^2}{4a} $$ Now we want to solve $y=0$ so $$ \begin{align} 0&=\frac{(2ax+b)^2}{4a}+c-\frac{b^2}{4a}\\ \frac{b^2}{4a}-c&=\frac{(2ax+b)^2}{4a}\\ b^2-4ac&=(2ax+b)^2 \end{align} $$ Solving for the positive and negative root yields the famous result.

Not the most efficient way to get this formula but it works.

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We can use calculus, at least, to motivate the choices made when completing the square: The axis of symmetry of the graph of a quadratic function $f(x) = ax^2 + bx + c$, $a \neq 0$, passes through its vertex, and in particular the $x$-value, $x_0$, of the line $\{x = x_0\}$ of symmetry is a critical point of the function: $$f'(x_0) = 2 a x_0 + b = 0.$$ Rearranging gives $$x_0 = -\frac{b}{2 a},$$ so in the coordinates $uy$, where $x = u + \frac{b}{2a}$, the axis of symmetry is the $y$-axis, $\{u = 0\}$. Substituting and simplifying gives that $$f(x) = a u^2 + \left(c - \frac{b^2}{4a^2}\right) ,$$ which we can solve for $u$ (and for $x$) simply by rearranging, compute the square roots of the expression for $u^2$, and rewriting in terms of $x$.

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