0
$\begingroup$

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?

$\endgroup$

3 Answers 3

1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $p\geq0$.

$\endgroup$
1
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.