# Solving this algebraic vector equation

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.

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

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

$$$$a^T = k_4 b^T - cb^TA$$$$

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.

$$$$x^T \left(k_1 I - 2 k_1 k_4 b b^T + 2 k_1 c b b^T A \right)x + 2 k_1 k_2 b^T x -k_3 || b ||^2$$$$

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.

• What are your ideas, what have you tried? Commented Mar 13, 2022 at 18:30
• So what I'm trying now is to isolate the $k_2 b \sqrt{...}$ term and 'square' both sides - by 'square' I'm going to try and basically take the 2-norm of both side. I'm not really sure what I'm aiming for though.. an equation in the form of $x^T A x + b x + c$ maybe?
– ABW
Commented Mar 13, 2022 at 18:49
• Not a bad idea. Please edit your question and show what you got. Also : does this equation have unique solution ? Commented Mar 13, 2022 at 19:31
• I'm not exactly sure how to show uniquness. I have a hunch the solution is exists and is unique but I don't know.
– ABW
Commented Mar 13, 2022 at 22:09
• Far from having a unique solution. Take the simplest quadratic forms of all: $x_1^2+x_2^2=r^2$. When $r>0$ its solutions are on a circle with radius $r$. Commented Mar 14, 2022 at 6:28

$$\def\l{\lambda} \def\L{\left}\def\R{\right} \def\LR#1{\L(#1\R)} \def\BR#1{\Big(#1\Big)} \def\c#1{\color{red}{#1}}$$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 $$x$$.
• How does one get to your rearranged equation $b k_5 = k_1 x$ and therefore $x= \lambda b$? What is $k_5$?
• @ABW The entire quantity in parentheses multiplying $b$ is a scalar value, I simply assigned it a convenient name, $\,k_5.\;$ The only vector quantities on each side of the equation are $b$ and $x$, so they must be scalar multiples of one another.
• I see now! Your solution makes perfect sense thanks so much. Furthermore one can add that because $k_5, k_1$ are strictly positive $\implies$ we must take the the positive solution from the quadratic equation in $\lambda$ which you derived.