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In my high school algebra class the teacher (who is me) says that the discriminant of a quadratic polynomial $ax^2 + bx + c$ is $b^2 - 4ac$.

I have read in the Wikipedia article that the discriminant of a polynomial is the product of the squares of the differences of its roots. This does not seem to be consistent with the above. If I subtract the roots of a quadratic and then square the result I get $\frac{(b^2 - 4ac)}{a^2}$.

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    $\begingroup$ What you read in the Wikipedia article holds for monic polynomials. $\endgroup$ – darij grinberg Dec 18 '15 at 17:35
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    $\begingroup$ The discriminant of an $n$'th degree polynomial is $\Delta=a_n^{2n-2}\prod_{i<j}(r_i-r_j)^2$, where $a_n$ is the leading coefficient and $r_1,r_2,\ldots, r_n$ are the roots. This is written in the discriminant Wikipedia article. In your case, it should be $a^{2\cdot 2-2}(r_1-r_2)^2$, which does equal $b^2-4ac$. $\endgroup$ – user236182 Dec 18 '15 at 17:40
  • $\begingroup$ Yes, thank you. I am reading the article more carefully now and it does clearly state that the discriminant is a_n^(2*n - 2) times the product of the squares of the differences of the roots where a_n is the leading coefficient. $\endgroup$ – Geoffrey Critzer Dec 18 '15 at 17:41
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Wolfram Mathworld says that the most common definition for the discriminant of a polynomial $p(z) = a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z +a_0$ is: $$D(p) \equiv a_n^{2n - 2} \prod_{i, j}^n(r_i - r_j)^2, i<j$$ This definition gives, and I quote,

...a homogenous polynomial of degree $2(n - 1)$ in the coefficients of $p$.

In the example of the quadratic equation $ax^2 + bx + c = 0$ (or $a_2 z^2 + a_1 z + a_0$), the discriminant is given by $b^2 - 4ac$ (or $D_2 = a_1^2 - 4a_2a_0$). You can extend this the the cubic equation $ax^3 + bx^2 + cx + d = 0$ (or $a_3 z^3 + a_2 z^2 + a_1 z + a_0 = 0$) and get $c^2 + b^2 - 4db^3 - 4c^3a + 18dcba - 27d^2a^2$ (or $D_3 = a_1^2 + a_2^2 -4a_0a_2^3 - 4a_1^3a_3 + 18a_0a_1a_2a_3 - 27a_0^2a_3^2$)

That is indeed the best explanation I can give.

Resources:

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