Let $p(x)$ be a $100$-degree polynomial with $100$ real and distinct roots, say $\alpha_1,\alpha_2,\cdots,\alpha_{100}$, and so $$p(x)=A(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_{100}),$$ where $A\in\mathbb{R}\setminus\{0\}$ and $α_{i}\neq 0$ for all $i\in[1,100]$. Find nature of roots of the equation
$$x^2p''(x)+3xp'(x)+p(x)=0$$
and also find nature of roots of the equation
$$10p(x)p''(x)=99(p'(x))^2.$$
Try:
$$(x^2p''(x)+2xp'(x)+xp'(x)+p(x)=0,\\ \frac{d}{dx}\bigg(x^2p'(x)\bigg)+\frac{d}{dx}\bigg(xp(x)\bigg)=0.$$
Could some help me to solve it, thanks in advance.
You have $$((px)'x)'=(px)''x+(px)'=(p''x+2p')x+(p'x+p)=p''x^2+3p'x+p.$$
The roots of $(px)'$ are the extrema of $px$, which are real and comprised in the $100$ intervals $(\alpha_k,\alpha_{k+1})$, where we define $\alpha_0:=0$.
Then again, the roots of $((px)'x)'$ are real and comprised in the $100$ intervals $(\beta_k,\beta_{k+1}$, where the $\beta_k$ are the above roots plus $\beta_0:=0$.
As an illustration, a simpler case with $\alpha_1=-1,\alpha_2=1,\alpha_3=2$ and the polynomials $\color{blue}p,\color{lightgreen}{(px)'},\color{magenta}{((px')x)'}$:
The second case can be solved from $(p^\alpha(x))''$.