# Problem concerning the real Nullstellensatz

The real quadratic Nullstellensatz says:

Let $p(x), q(x)\in \mathbb R[x_1,...,x_n]$ be quadratic polynomials such that $$(*) \ \{x\in \mathbb R^n: p(x)=0\}=\{x\in \mathbb R^n: q(x)=0\}$$ and $p(x)$ has at least one zero $x_0$ such that $grad \ p(x_0) \neq 0$. Then there is a $\alpha \in \mathbb R^*$ such that $p=\alpha q$.

Is the same true if the condition (*) is replaced by the following one: $$(*) \ \{x\in \mathbb R^n: p(x)=0\} \subset \{x\in \mathbb R^n: q(x)=0\} ?$$

According to my answer elsewhere to a related question, if $\{p=0\}$ is contained in $\{q=0\}$ then $p$ divides $q$ in $\mathbf R[x_1,\ldots,x_n]$. (If the condition of irreduciblity of $p$ there is not fulfilled, $p$ is simply the product of two linear polynomials that do not have a common zero in $\mathbf R^n$. That case is easy to handle.) Since $p$ and $q$ are both of degree $2$, there is $\beta\in\mathbf R$ such that $q=\beta p$. Since $q$ is not identically zero on $\mathbf R^n$, the real number $\beta$ is nonzero. Let $\alpha=\beta^{-1}$. Then $p=\alpha q$.