Converting from Quadratic to Second Order Cone optimization problem I want to convert the following problem into SOCP form:
$minimize \quad$ $x^TAx+a^Tx$:
$subject$ $to \quad$ $Bx \leq b$
The approach I am taking is introducing new variables, $u$ and $v$, such that:
$u=x^TAx$ and $v=a^Tx$.
Then, our problem becomes:
$minimize \quad$ $u+v$:
$subject$ $to \quad$ $Bx \leq b$, $u \leq x^TAx$ and $v \leq a^Tx$.
Among these 3 constraints, the first and the third are linear, so we can let them be as they are. Only the second constraint needs to be converted into 2-norm form somehow, which I am not able to get how to do. Could anyone kindly tell me how to do that conversion? Would I need to introduce more new variables, or does it involve using some formula or trick that I can't seem to think of?

This only works if $A$ is positive definite.  In this case, just drop the quadratic part and obtain
$$\begin{array}{rcl}
\min\limits_{x,y} && y + a^Tx\\
st && Bx\leq b\\
&& x^TAx \leq y
\end{array}$$
Then, we play a long circuitous game with this last constraint where
$$\begin{array}{rl}
&y\geq x^TAx\\
\Longrightarrow& 0\geq x^TAx - y\\
\Longrightarrow& 0\geq 4x^TAx - 4y\\
\Longrightarrow& 0\geq 4x^TAx + (1-y)^2-(1+y)^2\\
\Longrightarrow& (1+y)^2\geq 4x^TAx + (1-y)^2\\
\Longrightarrow& (1+y)^2\geq 4x^TAx + (1-y)^2, 1+y\geq 0\\
\end{array}$$
where the extra constraint, $1+y\geq 0$, can be added since $y\geq 0$ since $x^TAx\geq 0$ since $A\succ 0$ and $y\geq x^TAx$.  Anyway, then we have
$$\begin{array}{rl}
\Longrightarrow& 1+y\geq \sqrt{4x^TAx + (1-y)^2}, 1+y\geq 0\\
\Longrightarrow& 1+y\geq \sqrt{4x^TU^TUx + (1-y)^2}, 1+y\geq 0
\end{array}$$
where we use the Choleski factorization $A=U^TU$ since $A\succ 0$.  Then,
$$\begin{array}{rl}
\Longrightarrow& 1+y\geq \left\|\begin{bmatrix}2Ux\\1-y\end{bmatrix}\right\|, 1+y\geq 0\\
\Longrightarrow&  \begin{bmatrix}1+y\\2Ux\\1-y\end{bmatrix}\succeq_Q 0\\
\end{array}$$
Putting this all together, we have
$$\begin{array}{rcl}
\min\limits_{x,y} && y + a^Tx\\
st && Bx\leq b\\
&& \begin{bmatrix}1+y\\2Ux\\1-y\end{bmatrix}\succeq_Q 0
\end{array}$$
