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enter image description here 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.

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  • $\begingroup$ $P(P(x))=P(x)$ since $P(x)=x$ $\endgroup$ – Isham May 31 '18 at 2:09
  • $\begingroup$ Have you tried the discriminant definition of real roots? $\endgroup$ – Tony Hellmuth May 31 '18 at 2:09
  • $\begingroup$ You mean, (b-1)² - 4ac < 0 ?? I did try this, but it didn't get me anywhere $\endgroup$ – π times e May 31 '18 at 2:42
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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 $\mathbb{R}$.

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  • $\begingroup$ Thanks, finally I understood it. This was very intuitive. Thanks a lot :) $\endgroup$ – π times e May 31 '18 at 2:39
  • $\begingroup$ $\pi$ times $e$: You are wrong to thank this answer. If $P(x)\gt x$ then $ P (x) $ never cuts the first diagonal $ y = x $. This is not true when the minimum is $(a,b)$ with $a\gt b$. Your problem is less easy than you wonder. $\endgroup$ – Piquito May 31 '18 at 11:11
  • $\begingroup$ @Piquito What do you mean? If $P(x)>x$ or $P(x)<x$ is not true, then $P(x)=x$ must have real root! Where is the fault in my answer? Could you point directly? I think this problem has no realtion with the condition that the function is quadratic or not. The condition that $P(x)\in C^0(\mathbb{R})$ is enough! $\endgroup$ – W. mu May 31 '18 at 13:00
  • $\begingroup$ @W. mu: Try, dear friend, with $f(x)=x^2-6x+10$. This polynomial has no real roots but do you have $f(x)\le x$ over the interval $[2,5]$. $\endgroup$ – Piquito May 31 '18 at 15:28
  • $\begingroup$ @Piquito I don't care whether $f(x)=0$ has real roots, but whether $f(x)=x$ has real roots. $\endgroup$ – W. mu Jun 1 '18 at 0:35
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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.

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enter image description here 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

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    $\begingroup$ Because it depends just of $q$ and $r$, coefficients of the new equation. $\endgroup$ – Piquito May 31 '18 at 11:08
  • $\begingroup$ Which is why it says it's independent of alpha and beta, even though it has one of it's roots in terms of alpha. Makes sense, thanks a lot $\endgroup$ – π times e May 31 '18 at 14:54

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