$\mathbf{x^TDx}+\mathbf{b^Tx}=0\; \Rightarrow\;\exists!\left\{\mathbf{x}=0 \;\oplus\;\mathbf{x}=-\mathbf{D^-b}\right\}$?

Question

Consider $$\mathbf{x^TDx}+\mathbf{b^Tx}=0 \tag 1.$$

Let $\mathbf{D}$ is full rank diagonal matrix and $\mathbf{b} \neq 0$.

Then solutions $\mathbf{x}=0 \;\text{and}\;\mathbf{x}=-\mathbf{D^{-1}b}$ exist.

Question. Are there conditions that equation $(1)$ has another solutions?

Updated

Consider $$\left\{ \begin{align} \mathbf{y^TD_1y}+\mathbf{b_1^Ty}=0 \\ \mathbf{y^TD_2y}+\mathbf{b_2^Ty}=0 \end{align} \right. \tag 2$$

I believed that the system $(2)$ has a root if there are true conditions that an equation $(1)$ has another solutions, for example, $\mathbf{x}$ has infinite set of values.

Where did I make a mistake?

Updated

Really equation $(1)$ has another solutions, for example in $\mathbb{R^2}$, next root (not only)

$$ \mathbf{x}=\left[\frac{-b_1+\sqrt{-4d_1d_2x^2_2-4b_2d_1x_2+b_1^2}} {2d_1}\;,x_2\right]^\mathbf{T} $$

Answer 1

Pick any $x_0$ with $x_0^TDx_0\ne 0$ and $b^Tx_0\ne 0$ and consider the line $tx_0$. On this line $$ t^2x_0^TDx_0+tb^Tx_0=0\quad\Leftrightarrow\quad t=0\text{ or } t=-\frac{b^Tx_0}{x_0^TDx_0}\ne 0. $$ The second $t$ corresponds to a non-trivial solution $tx_0$, so there is a solution in almost all directions $x_0$.

Answer 2

If $D$ is diagonal and full rank, you know for sure that it is invertible, so such solution $\mathbb{x}=-D^{-1}b$ exists.