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The following is the quadratic formula derived by completing the square:

$$x = -\frac{b}{2a} \pm \sqrt{ \frac{b^2}{4a^2} - \frac{c}{a}} \tag{1}$$

This equation is equivalent:

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

We can obtain $(2)$ from $(1)$ by trying to simplify the term under the $\sqrt{\dots}$ as much as possible.

I've always felt there is no such thing as "the quadratic formula", and that both $(1)$ and $(2)$ are equally arbitrary.


Everything the discriminant $b^2 - 4ac$ can do, so can $\frac{b^2}{4a^2} - \frac{c}{a}$ (literally because it's just a scalar multiple of the other).

What makes the form $b^2 - 4ac$ so special? It does not seem to be "arbitrary"? As stated in the Wikipedia article for it:

The discriminant is widely used in polynomial factoring, number theory, and algebraic geometry

I could try to "formalize" the notion of:

  • "Maximally simplifying the term under the $\sqrt{\dots}$"
  • "The discriminant in $(1)$ normalized is the discriminant in $(2)$"

However this does not seem intuitive nor correct.


Why is the discriminant $b^2 - 4ac$ not arbitrary/special?

Does it appear elsewhere more "naturally"?

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  • $\begingroup$ If $a=0$ the regular definition still works :) $\endgroup$
    – Amit
    May 18, 2023 at 12:52
  • $\begingroup$ I can also write it as $x=\frac{-b}{2a} \pm \frac{\sqrt{4a^{2}b^{2}-16a^{3}c}}{4a^{2}}$, so we have $4a^{2}b^{2}-16a^{3}c$, which works for $a = 0$. $\endgroup$ May 18, 2023 at 16:00
  • $\begingroup$ Does it? I thought $0/0$ is a bit of a pickle $\endgroup$
    – Amit
    May 18, 2023 at 16:25
  • $\begingroup$ Ah sorry I see what you mean, but no, it won't work the same. It'll give you $0$ instead of $b$ for $a=0$ $\endgroup$
    – Amit
    May 18, 2023 at 16:29
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    $\begingroup$ I upvoted all answers. I want to point out it also follows from matrix theory and others. And there is also the lesser known formula with the square root in the denom ofcourse. $\endgroup$
    – mick
    May 18, 2023 at 19:21

3 Answers 3

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The discriminant generalizes to polynomials of any degree (not just quadratics). Given a polynomial $$p(x) =a_{n}\prod^{i}_{1\dots n} (x-r_{i}) $$ the discriminant is the formula for:

$$\Delta = a_{n}^{2n-2}\prod_{i<j}(r_{j}-r_{i})^2$$


For a second degree polynomial: $$p_2(x) = ax^2 + bx + c = \\ = a(x-r_{1})(x-r_{2})$$

the second degree discriminant formula gives: $$\Delta_{2} = b^2 - 4ac =\\ = a^2 \;(r_{2}-r_{1})^2 $$


For a third degree polynomial: $$p_3(x) = ax^3 + bx^2 + cx + d = \\ a(x-r_{1})(x-r_{2})(x-r_{3}) $$

the third degree discriminant formula is: $$\Delta_{3} = 18abcd – 4b^3d + b^2c^2 – 4ac^3 – 27a^2d^2 = \\ = a^4 \;(r_{2}-r_{1})^2(r_{3}-r_{1})^2(r_{3}-r_{2})^2 $$


The discriminant gives the (squared) product of the differences of distinct roots of a $n$'th degree polynomial, times the top coefficient to the power $2n-2$.

There exists formulas for higher degrees. The general formula for the discriminant is the determinant of the Sylvester matrix of $p(x)$ and $p'(x)$, divided by the top coefficient of $p$.


The discriminant formula can be characterized by the following properties:

  1. It's an algebraic formula in the coefficients of our polynomial. That is, it uses only sums and products of the coefficients, no fractions.

  2. It is equal to $0$ if and only if the polynomial has repeated roots. That is, it discriminates the polynomials which don't have $n$ distinct complex roots. This can be seen by looking at its expression in terms of the polynomial roots.

  3. It is the simplest formula with the above properties. Specifically, it is a divisor for any formula satisfying them. This makes it unique up to a constant factor.

More info on the generalized discriminant formula can be found here: https://en.wikipedia.org/wiki/Discriminant

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    $\begingroup$ A small typo: you want $\Delta = a_{n}^{2n-2}\prod_{i<j}(r_{j}-r_{i})^2$ (rather than $r_1$) $\endgroup$
    – Silverfish
    Oct 15, 2023 at 15:56
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The definition of the quadratic discriminant as $\,b^2 - 4ac\,$ makes it consistent with the definition of the discriminant for higher degree polynomials.

In general, for a polynomial $\,p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x +a_0\,$ of degree $n$ with roots $\,x_1, x_2, \ldots.x_n\,$ the discriminant can be defined as $\,\Delta = a_n^{2n-2} \prod_{1 \le i \lt j \le n}(x_i-x_j)^2\,$. For the quadratic $\,ax^2+bx+c\,$ this reduces to $\,\Delta = a^2(x_1-x_2)^2 = b^2 - 4ac\,$.

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Geometrically for a parabola with equation $y=ax^2+bx+c$ the coordinates of the vertex are given by $V=(x_v,y_v)=\left(-\frac b{2a},-\frac {\Delta}{4a}\right)$ such that the condition on the sign of $\Delta$ with the sign of $a$ determine the number of solutions for the quadratic equation $ax^2+bx+c=0$.

Moreover $\Delta$ appears in all the key parameter for the parabola as the focus $F=(x_f,y_f)=\left(-\frac b{2a},-\frac {1-\Delta}{4a}\right)$ and the related diretrix with equation $y=-\frac {\Delta}{4a}-\frac {1}{4a}$.


In multivariable calculus, in the simpler case of $2$ variables, the study of the nature of stationary points can be done with reference to the quadratic form

$$f(x,y)=ax^2+bxy+cy^2= \begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}a&\frac b2\\\frac b2&c\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$$

and of the Hessian determinant $$H=\begin{bmatrix}a&\frac b2\\\frac b2&c\end{bmatrix}=ac-\frac{b^2}4=-\frac{\Delta}4$$

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