Let $A,B$ be two ternary quadratic forms with real coefficients, given by symmetric matrices

$$\displaystyle 2A = \begin{pmatrix} 2a_{11} & a_{12} & a_{13} \\ a_{12} & 2a_{22} & a_{23} \\ a_{13} & a_{23} & 2a_{33} \end{pmatrix}, 2B = \begin{pmatrix} 2b_{11} & b_{12} & b_{13} \\ b_{12} & 2b_{22} & b_{23} \\ b_{13} & b_{23} & 2b_{33} \end{pmatrix}.$$

Let $V_\mathbb{R}$ denote the 12 dimensional real vector space of $(A,B)$ over $\mathbb{R}$, and let $G(\mathbb{R}) = \operatorname{GL}_2(\mathbb{R}) \times \operatorname{SL}_3(\mathbb{R})$. Let $(r,g)$ be an element of $G(\mathbb{R})$, where

$$\displaystyle r = \begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix}.$$

Then $(r,g)$ acts on $(A,B) \in V_\mathbb{R}$ by sending $(A,B)$ to $$(r_1 (gAg^T) + r_2 (gBg^T), r_3 (gAg^T) + r_4 (gBg^T)).$$

Bhargava, in his paper *The density of discriminants of quartic rings and fields*, stated that the stabilizer in $G(\mathbb{R})$ of an element $(A,B) \in V_\mathbb{R}$ has order 24 if $A,B$ have four common real zeroes over $\mathbb{P}^2$, order 8 if they have exactly one pair of real zeroes over $\mathbb{P}^2$, and order 4 if they have no common real zeroes. He simply said that "one easily checks" that this is the case. Can anyone give an explanation as to why this should be easy to see, and a proof of why it's true?