Stuck on this problem based on theory of equations/Quadratic equations

Question

I need help with this question. I'm stuck on it. I'll show how I approached this question, but I couldn't get very far.

$P(x) = x$ has no real roots

=> $P(x) - x$ has no real roots

=> $ax^2 - (b-1)x + c$ has no real roots

=>$ P(x) - x \gt 0$ or $P(x) - x \lt 0$ for all real values of $x$

Now I don't understand how to proceed further. Please help me out.

Answer 1

If $P(x)=x$ have no root on $\mathbb{R}$, then we assume $P(x)>x$ for $x\in\mathbb{R}$.($<$ is similar)

Since $P(P(x))>P(x)>x$, $P(P(x))=x$ has no root on $\mathbb{R}$.

Answer 2

Hint.-$$P(x)-x=0\iff Q_1(x)=ax^2+(b-1)x+c=0\\P(P(x))-x=0\iff Q_2(x)=(a^2+ab)x^2+ab+b^2-1)x+c(a+b+1)=0$$ What really matters are the polynomials $Q_1$ and $Q_2$. Making both discriminants negative is long and somewhat difficult. It is preferable to put $a\gt 0$ and establish that the minimum is greater or less than zero according to the sign of the coefficient $(a^2+ab)$ of $x^2$ in $Q_2(x)$ (the case $a\lt 0$ is analogue).

$\boxed{a\gt0}$ Taking derivatives as usual we have the condition $$Q_1\left(\frac{1-b}{2a}\right)\gt 0\quad(*)$$ Now you have as implication of $(*)$

$$\begin{cases} Q_2\left(\dfrac{1-b^2-ab}{2(a^2+ab)}\right)\gt 0\quad\text{ if }\qquad a^2+ab\gt 0\\ Q_2\left(\dfrac{1-b^2-ab}{2(a^2+ab)}\right)\lt 0\quad\text{ if }\qquad a^2+ab\lt 0\end{cases}$$ It is tedious but go for the negative discriminant is harder for something impracticable. At least that's how I think.

Answer 3

Please check out illustration 2.78 I understand what's going on here, except for one thing. How is the new (t³+qt-r=0) equation formed is independent of alpha and beta, if one of it's roots is 2 times alpha? Thanks