Expressing a real quadratic as the dot product of complex vectors I've been messing around with a quadratic that popped out of a problem I'm working on, and noticed it has a property I haven't seen before. The form is as follows:
$$ ax^2 + bx + c = 0$$
where
$$a = r_{1}^{2} +  r_{2}^{2} -  r_{3}^{2}$$
$$b = 2\left( r_{1} t_2 +  r_{2} t_2 -  r_{3} t_3 \right)$$
$$c = t_{1}^{2} +  t_{2}^{2} -  t_{3}^{2}$$
If we now define two vectors
$$ \underline u = [r_1, r_2, i r_3] \quad \underline v = [t_1, t_2, i t_3]$$
The quadratic can be re-written as
$$(\underline u \cdot \underline u) x^2 + 2(\underline u \cdot \underline v) x + (\underline v \cdot \underline v) = 0$$
and then
$$(\underline u x + \underline v) \cdot (\underline u x + \underline v) = 0$$
Does the fact that the equation possesses this property tell us anything useful? Can it be exploited in any way? is it a common property?
Thanks!
 A: I think that your question about the interest of relationship
$$\tag{1} (u.u)x^2+2(u.v)x+(v.v)=(ux+v).(ux+v)$$
has an answer in the representation of (1) by a quadratic form, an abstract concept that is rendered in finite dimensional vector spaces by matrix expressions of the form $X^TAX$. Here, the following transcription of (1) into quadratic form notation gives the easy-to-prove relationship :
$$\left(\begin{array}{cc}x& 1\end{array}\right)\left(\begin{array}{cc}u.u& u.v\\u.v&v.v\end{array}\right)\left(\begin{array}{c}x\\1\end{array}\right)=\left(\begin{array}{cc}x&1\end{array}\right)\left(\begin{array}{c}u^T\\v^T\end{array}\right)
\left(\begin{array}{cc}u&v\end{array}\right)\left(\begin{array}{cc}x\\1\end{array}\right)$$
that can as well be written using the versatility of matrix notations:
$$\left(\begin{array}{cc}x&1\end{array}\right)\left(\begin{array}{c}r_1&r_2&ir_3\\t_1&t_2&it_3\end{array}\right)
\left(\begin{array}{cc}r_1&t_1\\r_2&t_2\\ir_3&it_3\\\end{array}\right)\left(\begin{array}{cc}x\\1\end{array}\right).$$
A: Every quadratic equation can be written as
$$
0=\pmatrix{x& 1}\pmatrix{a&b/2\\b/2&c}\pmatrix{x\\1}
$$
The symmetric matrix in the middle can be decomposed as $LDL^T$ or some general $CSC^T$ form where $C$ is a diagonal matrix with $\pm 1$ entries on the diagonal.
Thus you can always write
$$
ax^2+bx+c=s_1(r_1x+t_1)^2+s_2(r_2x+t_2)^2,\ s_i=\pm 1
$$
so that your decomposition does not offer  any form of insight.
If all the squares (with real coefficients) get added, all $s_i=+1$, then you know that there are no real roots if at least one term is non-trivial, that is, has $t_i\ne 0$.
