Discriminant of quadratic function

Question

Prove that if we have two quadratic function $f(x)$ and $g(x)$ such that $|f(x)| >|g(x)|$, then $|\Delta_f|>|\Delta_g| $. I'm looking for hints, I've no idea where should I start. Thanks in advance...

Answer 1

NOTE I previously had up an erroneous "counterexample" But I now think the statement is true, and sketch here a proof.

Let the two quadratics be $p,q$ for which $|p(x)|>|q(x)|$ for all $x.$ Then we are trying to show $|\Delta_p|>|\Delta_q|$ for the two discriminants. Here $p$ can have no zeros, else if $p(r)=0$ then $0=|p(r)|>|q(r)|,$ not possible. Any replacement of $x$ by $x+h$, i.e. a shift of $x$ coordinates, has no effect on the hypothesis, and also has no effect on the discriminants, as may be checked. So we may assume things are centered so that the vertex of $p$ is at $x=0.$ So far we have the magnitude of $p(x),$ which we also call $p(x),$ is $kx^2+r,$ where $k,r>0$ [recall here p has no zeros and is positive.]

We would also like to scale each quadratic by multiplying by $m>0$ which changes $p$ to $mkx^2+mr$, and now set $(mk)(mr)=1$ to get $m^2=1/(kr),$ solvable with $m>0$ since $k,r>0.$ Using this multiplier $m$ we now have $p$ in the form $p(x)=ax^2+1/a.$ This rescaling by $m$ does not change inequalities on magnitudes of $p,q$ nor inequalities on their discriminants.

The usual discriminant of $ax^2+2bx+c$ is $4(b^2-ac),$ but from here we'll drop the factor $4$ and so use $b^2-ac$ for discriminants (keeping in mind to have the $x$ coefficient as $2b$) This said, the discriminant of $p$ is now $\Delta_p=-1.$ [I just find the extra 4 annoying.] Again this dropping of the factor 4 has no effect on comparing discriminant magnitudes.

Now let $q(x)=b(x-h)^2+c$ which has discriminant $\Delta_q=-bc$ (independent of $h$). The magnitude of $p$ which is now $p$ itself, can never equal that of either $q$ or $-q,$ else if it does at some $x=r$ we would have $|p(r)|=|q(r)|$ against the hypothesis. We need not consider the case $c=0,$ since then $\Delta_q=0<|\Delta_p|=1$ as desired. So the discriminant of $q$ is nonzero, and we can separate into two cases. In the first (easier) case $\Delta_q<0$ the quadratic $q$ has no zeros, and so its magnitude (which we also call $q$) may be expressed using $b,c>0$ in $q(x)=b(x-h)^2+c.$ When multiplied out,

$$q(x)=bx^2-2bhx+(bh^2+c). \tag{1}$$ Here $|q(0)|=bh^2+c<|p(0)|=1/a.$ In particular we have, since $b>0$, that $c<1/a.$ Now if we consider things for large $x$ we can get also $b \le a.$ To see this, divide each of $p,q$ by $x^2$ to get from the assumption to $$a+\frac{1}{ax^2} >b-\frac{2bh}{x}+\frac{bh^2+c}{x^2.}.$$ As $x \to \infty$ this relation implies $a \ge b.$ We now have both $b\le a$ and $c<1/a$ and may conclude that $bc<1.$ Since here $|\Delta_q|=|-bc|=bc<1=|\Delta_p|$ this finishes the case where $q$ has no zeros.

In the remaining case $q$ has two zeros, and its discriminant $\Delta_q=-bc>0.$ From what has been said already, there cannot be any intersection between $p$ and $q$ or between $p$ and $-q.$ If we use (1) to compute $p-q$ the resulting discriminant is $$ abh^2+ac+b/a-bc-1,$$ while the discriminant of $p-(-q)=p+q$ is the similar looking

$$-(abh^2+ac+b/a)-bc-1.$$ For the nonintersections mentioned, each of these discriminants need to be negative. This implies that $$|abh^2+ac+b/a|<bc+1.$$ Then we have $bc+1>0,$ i.e. $-bc<1,$ which since in the present case $|\Delta_q|=-bc$ (which is positive for this case) gives what is required in this case wherein $q$ has two zeros. (Note the single zero case has $\Delta_q=0$ already mentioned.