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.

I'm unable to find where to start with.

How do I mathematically prove this? I can visualise the scenario using a graph but, I don't get where to begin, mathematically.

Can someone help me out to proceed? Thanks in advance.

  • 2
    $\begingroup$ For starters, scaling by a constant preserves the roots but the scaled and original version satisfy the inequality. $\endgroup$ Aug 20, 2021 at 7:11
  • 1
    $\begingroup$ In fact ,whenever the polynomial on the rhs is zero,so is the polynomial on the lhs. This will immediately solve the situation when the rhs polynomial has two distinct roots. $\endgroup$ Aug 20, 2021 at 7:14

1 Answer 1


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.


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