I have to solve the following question:
In $\mathbb R^3$ with coordinates $a,b,c,$ sketch the surface (the discriminant) whose points represent polynomials $x^4 + ax^2 + bx + c$ with multiple roots.
As a hint to the question, we are also given the following:
Examine the curve formed by the polynomials $(x − t)^3(x + 3t)$ having a triple root, and show that the discriminant consists of the lines tangent to this curve.
I can't even begin to understand what the question is asking for, how to begin, or even if it's feasible to draw/sketch such a surface. I'd appreciate any help at this point.
Edit: Sorry for the lack of proper info, I genuinely do not know what the "discriminant" is referring to as well, but my best guess is the following https://en.wikipedia.org/wiki/Discriminant
- Comment 1: Cannot use any explicit discriminant formula.
- Comment 2: It's only asking us to sketch the surface.
- Definition: Multiple roots means the multiplicity of the root.
- Definition: discriminant of a polynomial is a quantity that depends on the coefficients and determines various properties of the roots.
- Definition: discriminant is either positive, zero, or negative.
- Definition: R^3 is the three dimensional space.
- Definition: Multiple roots consists of repeated two roots, triple roots, four roots, etc.
- Definition: Tangent lines is to a plane curve at a given point is the straight line that "just touches" the curve at that point.
An explicit expression of the discriminant as explained in the Wikipedia article you have browsed is under the form of the determinant of the following matrix, obtained by collecting the coefficients of
degree 4 polynomial $p(x)=x^4+0x^3+ax^2+bx+c$ and those of its
degree 3 derivative $p'(x)=4x^3+0x^2+2ax+b$,
and arranging them in successive rows with shifts, 3 times for the coefficients of $p(x)$ and $4$ times for those of $p'(x)$ (number of times = the degree of the other polynomial):
$$D=\begin{pmatrix} 1& 0& a& b& c& 0 &0\\ 0& 1& 0& a& b& c& 0\\ 0& 0& 1& 0& a& b& c\\ 4& 0& 2a& b& 0& 0& 0\\ 0& 4& 0& 2a& b& 0& 0\\ 0& 0& 4& 0& 2a& b& 0\\ 0& 0& 0& 4& 0& 2a& b\\ \end{pmatrix}\tag{1}$$
The 3D surface you are asked to find has equation $\det(D)=0$, i.e., using a Computer Algebra System
$$16a^4c - 4a^3b^2 - 128a^2c^2 + 144ab^2c - 27b^4 + 256c^3=0\tag{2}$$
If you take $b$ as the main variable, i.e., if you write equation (2) under the form:
$$-27b^4 + 4a(36c-a^2)b^2 + 16c(a^4 - 8a^2c + 16c^2)=0.$$
Setting $B=b^2$, one recognizes a quadratic equation:
$$-27B^2 + 4a(36c-a^2)B + 16c(a^4 - 8a^2c + 16c^2)=0\tag{3}$$
Solving (3) gives 2 roots:
$$B_1,B_2=\dfrac{2}{27}((36ac-a^3) \pm (a^2+12c)^{3/2})$$
In a second step, one comes back return to variable $b$, getting finally 4 expressions :
$$b=\sqrt{B_1}, \ \ b=-\sqrt{B_1}, \ \ b=\sqrt{B_2}, \ \ b=-\sqrt{B_2}$$
each one being an equation of a part of the surface you are asked to produce.
Edit: here is this (self intersecting) surface (caution: the vertical plane $b=0$ is spurious; it is an artefact of construction):
Fig. 1: Graphical representation of equation (2) with semi-transparency: a so-called "swallow-tail". Direction of increasing $a,b,c$ is resp. NW-SE,SW-NE and South-North.
Here is the Matlab program I have written for its generation :
Remark : I just realized that the figure above was given as the second figure in the Wikipedia article.
Remark 1: An excellent reference for this kind of issues.
Remark 2: Expression (1) can be found as well here.
Remark 3: I am surprized you are assigned such a task, without having done progressive work on this subject. Besides, I am still asking myself how the polynomial $(x-t)^3(x+3t)$ can help with this question.