# Nature of roots of two quadratic expressions

Given $$\,\left|px^2 +qx +r\right|\leqslant\left|Px^2 +Qx +R\right|\,$$ for all real $$x$$ where $$P,Q$$ and $$R$$ are different from $$p,q$$ and $$r$$, I wish to find the relation between the roots of these quadratic expressions assuming both of them have real roots.So, their discriminants are both positive.

Let us denote by $$g(x)= px^2+qx+r$$, and $$f(x)= Px^2+ Qx+ R$$. We have $$|g(x)|\leq |f(x)|.$$ They are real quadratic polynomials with non-negative determinants. If $$a,b$$ are the roots of $$f$$, then $$|g(a)|\leq |f(a)|=0,$$ i.e., $$g(a)=0$$, and similarly $$g(b)=0$$. So, if $$a\neq b$$ then $$a,b$$ are also the two roots of $$g$$. We may also assume $$P\neq 0$$, because if $$P=0$$, then $$p=0$$ by taking $$x\to \infty$$, otherwise we have two linear polynomials with the same root by our previous discussion. If $$a=b$$, then we may write $$f(x)= P(x-a)^2,$$ for $$a= -\frac{Q}{2P}$$. Since $$a$$ is also a root of $$g$$ we may factor it into $$g(x)= p(x-a)(x-c).$$ Divide by $$|g(x)|\leq |f(x)|$$ by $$|x-a|$$ to get $$|p(x-c)|\leq |P(x-a)|,$$ and plug-in $$x=a$$ to get $$c=a$$. As a result, in any case, the roots of $$f,g$$ are the same.