Prove that two coefficients of a quadratic function are always positive

Question

I found an interesting question from a Chinese maths textbook but did not obtain the full results stated.

Question: Let $f(x)$ be a quadratic function of the form $x^2+px+q$ such that the equation $f(f(x))$ $=0$ has two equal real roots. Show that $p$ $\geqslant$ $0$ and $q$ $\geqslant$ $0$.

It is easy to show that $q$ $\geqslant$ $0$ by using the discriminant $b^2-4ac=0$ for the equation $f(f(x))$ $=0$. However, the proof that $p$ $\geqslant$ $0$ is still unknown. Can somebody help me with it?

Answer 1

Since $f(f(x))=0$ does have a solution, $f(x)$ can be factored, into say $f(x)=(x-\alpha)(x-\beta)$. Here, we have $\alpha+\beta=-p,\alpha\beta=q.$

If $\alpha=\beta,$ you can show $\alpha$ and $\beta$ must be $0$. In this case, we have $p=q=0.$

Now, assume $\alpha\neq \beta.$ Then, $f(f(x))=(f(x)-\alpha)(f(x)-\beta)=0$ is equivalent to $f(x)-\alpha=0$ or $f(x)-\beta=0.$ Since both of these equations cannot be true at the same time (since $\alpha\neq \beta$), and $f(f(x))=0$ has only one solution, it must be the case that one of the two equations has only one root (i.e., a double root), and the other no roots.

Without loss of generality, we can assume $f(x)-\alpha=0$ has a double root, and $f(x)-\beta=0$ has no roots. Looking at the discriminant, we get $(-\alpha-\beta)^2-4(\alpha\beta-\alpha)=0$ and $(-\alpha-\beta)^2-4(\alpha\beta-\beta)<0.$

Now, we have $(-\alpha-\beta)^2-4(\alpha\beta-\alpha)=(\alpha-\beta)^2+4\alpha=0,$ so $\alpha\le 0.$ Similarly, we have $\beta<0.$ Hence, $p=-\alpha-\beta\ge 0,$ and $q=\alpha\beta\ge 0.$

Answer 2

Note that for every $x$ you have $f(x)=f(-p-x)$ so if $f(f(x))=0$ then also $f(f(-p-x))=0$. Hence if $f(f(x))=0$ has only one real root $y$, assuming that $p$ is real, we have $y=-p-y$ and hence $y=-\tfrac{p}{2}$.

Now plugging this in yields $f(y)=q-\tfrac{p^2}{4}$ and hence $$0=f(f(-\tfrac p2))=(q-\tfrac{p^2}{4})^2+p(q-\tfrac{p^2}{4})+q=\tfrac{p^4}{16}-\tfrac{p^3}{4}-q\tfrac{p^2}{2}+pq+q^2,$$ or equivalently $$(4q)^2+2p(2-p)(4q)-4p^3+p^4=0,$$ and hence by the quadratic formula $$4q=p(p-2)\pm2p,$$ which shows that $4q=p^2$ or $4q=p(p-4)$. In the latter case $f(y)=-p$ and $$f(f(y))=(-p)^2+p(-p)+q=q,$$ which shows that $q=0$ and hence either $p=0$ or $p=4$. In either case $p\geq0$.

Answer 3

There are two cases: first when f(x) has 1 zero, second when f(x) has 2 zeros.

1. f(x) has 1 zero

Let the zero a

f(x)=(x-a)^2=x^2-2ax+a^2

f(f(x))=0 has 1 root, so f(x)=a has 1 root.

x^2-2ax+a^2-a=0

D=4a^2-4(a^2-a)=4a

discriminant needs to be 0, so a=0

then p=-2a=0 , q=a^2=0

2. f(x) has 2 zeros

Let two zeros a and b

f(x)=(x-a)(x-b)=x^2-(a+b)x+ab

f(f(x))=0 has 1 zero, so there is total 1 root for

x^2-(a+b)x+ab-b=0 and x^2-(a+b)x+ab-b=0

WLOG x^2-(a+b)x+ab-a=0 has 1 zero and x^2-(a+b)x+ab-b=0 has no zero

D_1=(a+b)^2-4(ab-a)=(a-b)^2+4a=0 , D_2=(a+b)^2-4(ab-b)=(a-b)^2+4b<0

since (a-b)^2>=0 , a<=0 and b<=0

then p=-(a+b)>=0 , q=ab>=0

To conclude, p>=0 and q>=0

BTW the conclusion for #2 includes #1

(I recognized later)

This is my first time writing my answer :)

sorry if it was hard to read