Find the minimum of the following quadratic function
$U(x) = \frac{1}{2}x'Qx -qx'R +\frac{\alpha}{2}(x'd -b)^2$
Where Q is a positive semi-definite symmetric matrix, $x,R,d$ are $Nx1$ vectors, and $q, \alpha$ and $b$ are scalars $>0$.
The problem is a penalty approximated solution to
$U(x) = \frac{1}{2}x'Qx -qx'R$
Such that $x'd=b$.
The solution is to set the gradient equal to 0, which I do here:
$Qx - qR +\alpha (x'd-b)d = 0$
It is here where I get stuck. I am not sure how to pull the x out of $(x'd-b)d$.
Also, I'm assuming that I will be able to invert Q, or the matrix that becomes Q, to solve the program as the matrix is positive semi-definite and therefore is convex. Is this thinking on the right path?
A very similar question is asked here, but with b = 0: Finding minimum of penalty-approximated quadratic problem