Solution of quadratic equation with some constraints Setting the following constraints on $(A, B,C, D)$ (reals):
$ A +  \frac{B}{2} + \frac{C}{4} +  \frac{D}{8}=0$
$ A +  \frac{B}{4} + \frac{C}{16} +  \frac{D}{64}=0$
Is it always possible to find $(A, B,C, D)$ such that:
$$ (3 AB +\frac{12}{8} BC  + \frac{48}{64} CD) X_1 + (63 A D + \frac{60}{8} BD) X_2 + 15 A C  X_3 =0$$
Where $X_1, X_2, X_3$ are fixed real numbers ?
I suppose there is except if the associated quadratic form is definite positive, but how to characterise a definite positive form ? Any reference ? 
 A: From the constraints, you can get that:
$$\begin{cases}
A = \displaystyle\frac{C}{8} + \frac{3D}{32}\\
B = \displaystyle-\frac{3C}{4} - \frac{7D}{16}
\end{cases}.$$
Pluggin this in the 3rd equation, you get:
$$C^2 (960X_3 - 720X_1) + D^2(1344X_2 - 63X_1) + CD(720X_3 + 1152 X_2 - 144 X_1) = 0.$$
Starting from this point, you can write your equation as a quadratic form in matricial fashion:
$$v^\top M v = 0,$$
where
$$v = \begin{bmatrix}C\\D\end{bmatrix} ~\text{and} ~ M = \begin{bmatrix}960X_3 - 720X_1 & \displaystyle\frac{720X_3 + 1152 X_2 - 144 X_1}{2}\\\displaystyle\frac{720X_3 + 1152 X_2 - 144 X_1}{2} & 1344X_2 - 63X_1\end{bmatrix}.$$
Now, you need to look to the eigenvalues of $M$ in order to say that $v^\top M v$ is a definite positive quadratic form. For reference on this, please take a look here.
Anyway, this approach can be a little bit messy, since the matrix $M$ depends on $3$ real parameters $X_1$, $X_2$ and $X_3$.
Instead, I would start by looking if $M$ is not full rank. In that case, then there exists a $v \neq [0, 0]^\top$, such that 
$$v^\top M v = 0.$$
