Which is the correct way/ method/Formula to find the discriminant of a quadratic equation $f$ in 3 variables?
i.e., a quadratic form in 3 variables. Also, how to conclude that, whether this $f$ is reducible or irreducible?
Some one knows this please help. Thank you very much
Let us write the general 2nd degree equation under the form:
$$\tag{1}ax^2+2bxy+cy^2+2dxz+2eyz+fz^2.$$
In order for (1) to be expressible under the form of a product of two first degree polynomials:
$$\tag{2}(px+qy+rz)(sx+ty+uz),$$
coefficients $a,b,c,d,e,f$ must satisfy the pair of constraints :
$\tag{3}\begin{cases}b^2 + d^2 + e^2 - ac - af - cf & = & 0\\ acf + 2bde - cd^2 - fb^2 - ae^2 & = & 0\end{cases} \ \ \ \text{with} \ \ \ a + c + f \neq 0.$
Here is a linear algebra explanation.
Let us write equation (1) under its classical matrix form :
$$\begin{pmatrix}x&y&z\end{pmatrix}\underbrace{\begin{pmatrix}a&b&d\\b&c&e\\d&e&f\end{pmatrix}}_A\begin{pmatrix}x\\y\\z\end{pmatrix}$$
Relationship (2) is known to be true (see Appendix) if and only if matrix $A$ has rank one, i.e., if and only if it has exactly two zero eigenvalues.
Expressing that $A$ has exactly 2 zero eigenvalues is the same as saying that its characteristic polynomial is equal to $-\lambda^3+k\lambda^2$ for some nonzero $k$.
Identifying $-\lambda^3+k\lambda^2$ to the characteristic polynomial of $A$:
$$\det(A-\lambda I_3)= - \lambda^3 + (a + c + f) \lambda^2 +(b^2 + d^2 + e^2 - ac - af -cf)\lambda + (acf + 2bde - cd^2 - fb^2 - ae^2),$$
we obtain conditions (3).
Appendix : Rank 1 property is due to the fact that (2) can expressed under the following form:
$$\begin{pmatrix}x&y&z\end{pmatrix}\begin{pmatrix}p&&\\&q&\\&&r\end{pmatrix}\begin{pmatrix}0&&\\&0&\\&&1\end{pmatrix}\begin{pmatrix}s&&\\&t&\\&&u\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}.$$
Remark : A quadratic form in three variables set equal to 0 is the equation of a quadric surface passing through the origin. For example :
$$x^2+y^2-z^2=0 \ \ \text{is the equation of a circular cone with axis Oz.}$$
Expression (2), set equal to $0$, instead of a surface, corresponds to the degenerated case of 2 planes passing through the origin.
Remark: Let
$$\underbrace{\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}}_U, \ \ \underbrace{\begin{pmatrix}1&0&1\\-1&1&1\\0&-1&1\end{pmatrix}}_P, \ \ \underbrace{\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}}_D \ \ $$
Rank-one matrix $U$ could have taken instead of rank-one matrix $D$ because
$$\tfrac13U=PDP^{-1}$$
(Columns of matrix $P$ are eigenvectors of $U$ resp. associated with eigenvalues $0,0,1$).
