Solve $x \sqrt{x^T A x} + c =0$ for x in terms of $A$ and $c$ only

Question

Question: Solve $x \sqrt{x^T A x} + c =0$ for x in terms of $A$ and $c$ only.

Here $x \in \mathbb{R}^n$, $c \in \mathbb{R}^n$, and $A \in \mathbb{R}^{n\times n}$. Lastly, $A$ is symmetric and PSD, not sure if that matters.

This seems like it should be doable, but I am struggling to isolate $x$ such that no components of $x$ show up on the right hand side.

I also have tried this for the case where $A = I$, but still no luck.

Answer 1

There are two possibilities:

1. $c=0$. The equation then becomes $x\sqrt{x^TAx}=0$. Hence $x=0$ or $x^TAx=0$. In both cases, $x^TAx=0$ is true. Therefore the solution set is $\ker(A)$.
2. $c\ne0$. The equation thus implies that $x$ is a nonzero scalar multiple of $c$. Let $x=kc$ for some $k\in\mathbb R$. The equation then becomes $(k|k|\sqrt{c^TAc})c=-c$. Hence it is solvable only if $Ac\ne0$. Suppose this is the case. Then $k|k|=-\frac{1}{\sqrt{c^TAc}}$. Therefore $k$ must be negative. In turn, $k=-(c^TAc)^{-1/4}$ and the equation has a unique solution $x=-(c^TAc)^{-1/4}c$.