Solving a system of equations for x that contains $x^Tx$ (where x is a vector) So I may as well be a complete novice as far as math goes, but I need to solve a system of equations that contains some matrix multiplications. Sorry for lack of background knowledge / difficulty understanding. 
I have two variables, $L$ and $x$. $L$ is a scalar and 
$x$ is a column vector. There is a "known" constant $c$ that has the same dimensions as $x$. Additionally, I have a "known" square symmetric matrix $Q$. This matrix $Q$ is positive-definite, but I don't think that's relevant to solving the system of equations. 
So I need to solve for the two unknowns $L$ and $x$ in terms of the other values in the following equations:
$c^T - (2*L*x^TQ) = 0$
$1 = x^TQx$
So if $x$ and $Q$ were just scalars, this would be really simple. You could divide the second equation by $Q$ and take the square root, but is something analogous to that even possible with the matrix multiplications?
Sorry again for lack of knowledge. 
 A: The first equation is equivalent to $2LQ^T x = c$, or $Q^T x = \frac{1}{2L}c$. Since $Q$ is symmetric, this simplifies to $Qx = \frac{1}{2L}c$. Assuming $Q$ is strictly positive definite ($x^T Q x > 0$ for all nonzero $x$), it follows that $Q$ is invertible. Therefore, $x = \frac{1}{2L}Q^{-1}c$ is the unique solution to the first equation. 
Plugging this solution into the second equation gives us
$$\begin{aligned}
x^T Q x
&= \left(\frac{1}{2L}Q^{-1}c\right)^T Q \left(\frac{1}{2L}Q^{-1}c\right) \\
&= \frac{1}{4L^2}c^T Q^{-1} Q Q^{-1}c \\
&= \frac{1}{4L^2}c^T Q^{-1} c \\
\end{aligned}$$
where we have used the fact that the transpose of $Q^{-1}$ is still $Q^{-1}$, since $Q$ is symmetric.
We want to find the value of $L$ which makes this equal to $1$:
$$\frac{1}{4L^2}c^T Q^{-1} c = 1$$
Multiplying both sides by $L$ and taking the square root gives us
$$L = \pm\frac{1}{2}\sqrt{c^T Q^{-1} c}$$
Note that the square root on the right hand side is a real number since $c^T Q c > 0$ (because $Q$ is positive definite).
