Solving this algebraic vector equation

Question

How does one solve this equation for the vector $x$? The terms $b$, $a$ are nonzero vectors and $k_1, k_3, k_4>0$ are positive constants and $k_2$ is a real number. Note that $k_3$ makes the number under the square root positive. Also note that $\left(a^T x - k_2 + \sqrt{(a^T x - k_2)^2 + k_3} \right) > 0$ because $k_3$ is positive.

\begin{equation} 0 = -k_1 x + b \left(a^T x - k_2 + \sqrt{(a^T x - k_2)^2 + k_3} \right) \end{equation}

Also $A \succ 0$ is a positive definite matrix and $c>0$,

\begin{equation} a^T = k_4 b^T - cb^TA \end{equation}

Does anyone have any ideas?

Edits: I've added some more specifics to the problem.

What I've tried:

The first thing I notice is that if the component $b_i = 0$ this implies that $x_i$ will be zero uniquely. To try and solve for $x$, I've isolated the square-root term and taken the squared 2-norm of both sides. After some simplification, I get is the following.

\begin{equation} x^T \left(k_1 I - 2 k_1 k_4 b b^T + 2 k_1 c b b^T A \right)x + 2 k_1 k_2 b^T x -k_3 || b ||^2 \end{equation}

So now this is in the standard quadratic form. But now I've got to try and solve this for x.. I believe I should try completing the square? Would this resolve the issue with the nonsymmetric term $b b^T A$?

So I guess the next thing I will try is to complete the square, I need to brush up with some notes on this. Hopefully then I'll get a things in the form of $(x - p)^T H (x - p) = c$ where $H$ is positive definite. This should yield the a unique solution.

Answer 1

$ \def\l{\lambda} \def\L{\left}\def\R{\right} \def\LR#1{\L(#1\R)} \def\BR#1{\Big(#1\Big)} \def\c#1{\color{red}{#1}} $First, calculate the dot product of the two known vectors $$\eqalign{ \pi &= a^Tb \\ &= k_4 b^Tb - cb^TAb \\ &= b^T\big(k_4I-cA \big)b \\ }$$ The vector equation can be rearranged to $$\eqalign{ b\,(k_5) = k_1x \quad\implies\quad x=\l b \\ }$$ The only remaining task is to determine is the value of the scalar $\l$. $$\eqalign{ &\LR{\c{\pi\l} - \c{k_2} + \sqrt{(\pi\l - k_2)^2 + k_3} } = \c{k_1\l} \\ &\sqrt{(\pi\l - k_2)^2 + k_3} = {\c{(k_1-\pi)\l + k_2}} \\ &\LR{\sqrt{(\pi\l - k_2)^2 + k_3}}^2 = \BR{{(k_1-\pi)\l + k_2}}^2 \\ &{\pi^2\l^2 -2k_2\pi\l + k_2^2 + k_3} = (k_1^2-2k_1\pi+\pi^2)\l^2 + 2k_2(k_1-\pi)\l + k_2^2 \\ &0 = (k_1^2-2k_1\pi)\c{\l^2} + 2k_2(k_1)\c{\l} - k_3 \\ }$$ This quadratic equation will produce two solutions for $\l,$ and therefore two solutions for $x$.