# Prove that two coefficients of a quadratic function are always positive

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?

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 $$q=\alpha\beta\ge 0.$$

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 $$p\geq0$$.

• Anyone care to explain the downvote? Apr 6, 2020 at 10:21

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

1. 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