In the following, when I talk about the zero of a homogeneous polynomial I always mean a projective zero.
Let $ q $ be a real quadric. Then $ q $ has a real zero if and only if $ q $ has indefinite signature as a quadratic form (in other words, isotropic).
Let $ q_1 $ and $ q_2 $ be real quadrics. What is a sufficient condition for $ q_1,q_2 $ to have a real intersection? (in other words a shared real zero).
Note that while it is necessary for $ q_1 $ and $ q_2 $ to both be of indefinite signature that is not sufficient. For example $$ q_1(x_1,x_2,x_3)=x_1^2+3x_1x_2+x_1x_3+x_2x_3-x_3^2 $$ and $$ q_2(x_1,x_2,x_3)=-x_1x_2+x_2^2-x_1x_3-x_2x_3+x_3^2 $$ both have signature $ (2,1,0) $, but they have no real intersection. To see why, note that $ q_1+q_2=x_1^2+2x_1x_2+x_2^2=(x_1+x_2)^2 $. So at any shared 0 we have $ (x_1+x_2)^2=0 $. Thus at any shared real zero we have $ x_2=-x_1 $. Plugging this into $ q_1 $ we have that at any shared real zero $ -2x_1^2-x_3^2=0 $. So we must have $ x_1=0=x_3 $ which in turn implies that at any shared real zero $ x_2=-x_1=0 $. Thus $ q_1,q_2 $ have no real intersection.
Note on the complex case: Bezout's theorem says that if $ q_1,q_2 $ are quadrics in 3 or more variables then that is sufficient to guarantee a shared complex zero.
