I've been stuck on this problem for a while now :
Let $a\geq0$.
Show the following polynomial has exactly one positive root for all $k\geq 0$: where k is a real
$$f(a)=k^2a^4 + 2ka^3 -(k^3+k^2-1)a^2-2k^2a-k$$
More over prove that the corresponding root $a_0 \in \mathopen[\sqrt{k-1},\sqrt{k}\mathclose]$ for $k\geq 1$
My take:
First of all we can easily prove there exists a positive root $\alpha$ for $f(a)$ using IVT.
Then i computed the first and second derivative:
$$f'(a)=4k^2a^3+6ka^2-2(k^3+k^2-1)a-2k^2$$
$$f''(a)=12k^2a^2+12ak-2(k^3+k^2-1)$$
also $f''(a)$ is increasing over all the positive reals because $12k^2\geq 0$ and $12k\geq0$
$$f''(a)=0\Leftrightarrow a=-\frac{1}{2k}+\frac{\sqrt{3(3k^2+3k^3+1)}}{6k}$$
but $$a \geq 0 \Leftrightarrow 3k^3+k^2-2\geq 0 \Leftrightarrow k\geq 0.755$$
so if $k < 0.755$ then $f''(a)>0$ for all $a\geq0$ which means $f'(a)$ is strictly increasing over the positive the reals.
Also since $$f'(0)=-2k^2\le 0$$and $$\lim_{a\to\infty} f'(a)=+\infty$$
then by the intermediate value theorem there exists a positive real number $a_0$ such that $f'(a_0)=0$
Case 1:
$\alpha \in \mathopen[0,a_0\mathclose]$
Suppose there exists another positive real $\beta$ such that $f(\beta)=0$
Then by Rolle's theorem there exists a $c\in \mathopen]\alpha,\beta[$ such that $f'(c)=0$
Case 1-a
if $\beta \in \mathopen[0,a_0\mathclose[$ then c is in the same interval so by montony of $f'$ we get $0=f'(c)<f'(a_0)=0$ which is a contradiction
Case 1-b
if $\beta \in \mathopen[a_0,+\infty\mathclose[$ then $c\in \mathopen[0,+\infty\mathclose[$
but here it's possible to find $f'(c)=0$ for $c=a_0$ since $a_0$ lies in $\mathopen[0,+\infty\mathclose[$
While plotting it in Desmos it looks like for all $k\ge 0$ we have $a_0<\alpha$ which would solve the problem since for the second case where $\alpha \in \mathopen[a_0,+\infty\mathclose[$,
Rolle's theorem applies directly but i can't prove it , and even then i have to prove the same proposition for $k\geq 0.755$ which i am unable to do .
Thanks in advance for any help.
Best Answer
To prove that a positive root exists and is unique:
For any $k>0,$ there is exactly one sign change in the sequence of polynomial's coefficients:
$$f(a)=\underbrace{k^2}_+a^4 \underbrace{+ 2k}_+a^3 \underbrace{-(k^3+k^2-1)}_\pm a^2\underbrace{-2k^2}_-a\underbrace{-k}_- $$ By Descarte's rule of signs, the polynomial has exactly one positive real root for any $k>0.$
I have doubts that the root lies in $[\sqrt{k-1},\sqrt{k}],$ because $$f(\sqrt{k})=-k^3<0,$$ but $$\lim\limits_{a\to\infty}f(a)=\infty.$$