Question

Linear Algebra question. Show that the two polynomials ax^2 + bx + c = 0 and dx^2 + ex + f = 0, where a not = 0 and d not = 0 have a common root if and only if the following determinant equals 0. The determinant is 4 by 4 with the 1st row: a b c 0, 2nd row: 0 a b c, 3rd row: d e f 0, 4th row: 0 d e f. If this will help, the determinant equation becomes: aaff + acee - acdf - abef - bcde + bbdf + ccdd - acdf = 0. So, somehow, I have to relate the determinant equation expansion to (-b +-sqrt(b^2 - 4ac)) / 2a = (-e +-sqrt(e^2 - 4df)) / 2d, but I don't see this last equation coming down to the determinant expansion. I think I'm on the right track, but I'm still missing something key.

Answer 1

Suppose the two polynomials do have a common root then
\[ \large ax^2+bx+c=dx^2+ex+f \]
then
\[ \large (a-d)x^2+(b-e)x+(c-f)=0 \]
has a root.

Answer 2

That is true as far as it goes, but that will only find the root(s). It still does not show any relationship between the 6 variables and the determinant.

Answer 3

since it has a root then
\[ \large (b-e)^2-4(a-d)(c-f)\geq0 \]

Answer 4

As you probably know, the term determinant has more than one meaning. I am not referring to the term b^2 - 4ac in the general solution to a quadratic equation. I am referring to a 4x4 array of elements, similar to the mathematical definition of a matrix.

Answer 5

on the other hand
\[ \large \begin{vmatrix}
a & b & c & 0\\ 0 & a & b & c\\ d & e & f & 0\\ 0 & d & e & f
\end{vmatrix}=
a(af^2+e^2c-cdf-bef)+d(b^2f+c^2d-bce-acf) \]

Answer 6

If this will help, the determinant equation becomes: aaff + acee - acdf - abef - bcde + bbdf + ccdd - acdf = 0.

Answer 7

@satellite73

Answer 8

For anyone interested, I solved it. The last equation, after simplifying it, does indeed equal the determinant expansion equation. It appears that I had it figured out all along but just did not realize it.