Reciprocated Insane Roots

By transformation of roots, $x = \frac {1}{\alpha^2}$

So $\alpha=\frac {1}{√x}$

$0= 7x^2√x + 6x^2 + 16x√x + 7x + 9√x + 1$

$[(7x^2 + 16x + 9)√x]^2 = [-(6x^2 + 7x + 1)]^2$

Considering only terms of $x^5\quad and\quad x^4$

$49x^5 + 224x^4 + ... = 36x^4 + ...$

So sum of roots $=-\frac {188}{49}$

Answer = 139

$By~Vieta.... Sum~of~ reciprocal ~of~ roots~S_+=- \dfrac 6 7 .\\Sum~of~ reciprocal ~of~ roots~multiplied~ two~ at~ a~time~S_X=\dfrac{16} 7 .\\ S_{+^2}=Sum~of~sqares~of~reciprocal~of~the~roots.\\\therefore~S_+^2=S_{+^2} +2*S_X,\\\implies~\left(- \dfrac 6 7 \right)^{\large 2}=S_{+^2} +2*\dfrac{16} 7\\ \therefore~S_{+^2}=\dfrac{36}{49}- \dfrac{32} 7 \\\implies~~S_{+^2}=- \dfrac {188}{49} = -\dfrac p q . ~~~~~\implies~p-q =~~~~~~~~~~~~~\color{#D61F06}{ \large 139}$

$\begin{aligned} f(x) &=\prod\limits_{i=1}^5 (x - x_i) \text{ where } x_i \text{ are the roots of } f(x) \\ \ln(f(x)) &= \ln\left ( \prod\limits_{i=1}^5 (x - x_i) \right ) = \sum\limits_{i=1}^5 \ln(x - x_i) \\ \frac{f^{'}(x)}{f(x)} &= \sum\limits_{i=1}^5 \frac{1}{x - x_i} && \color{#D61F06} \text{differentiate both sides}\\ \frac{f^{''}(x)f(x)-\left ( f^{'}(x) \right )^2}{f^2(x)} &= \sum\limits_{i=1}^5 \frac{-1}{(x - x_i)^2} && \color{#D61F06} \text{again differentiate both sides}\\ \frac{f^{''}(0)f(0)-\left ( f^{'}(0) \right )^2}{f^2(0)} &= \sum\limits_{i=1}^5 \frac{-1}{(0 - x_i)^2} =\sum\limits_{i=1}^5 \frac{-1}{x^2_i} && \color{#D61F06} \text{substitute x = 0}\\ \sum\limits_{i=1}^5 \frac{1}{x^2_i} &= - \frac{188}{49} && \color{#D61F06} \text{see the Note}\\ \color{#3D99F6} Note: \\ &\color{#3D99F6} f(x)= \sum\limits_{k=0}^5 a_k x^k \Rightarrow f(0) = a_0 = 7\\ &\color{#3D99F6} f'(x)= \sum\limits_{k=1}^5 k \cdot a_k x^{k-1} \Rightarrow f'(0) = 1 \cdot a_1 = 6 \\ &\color{#3D99F6} f''(x)= \sum\limits_{k=2}^5 k(k-1) \cdot a_k x^{k-2} \Rightarrow f''(0) = 2 \cdot 1 \cdot a_2 = 2 \cdot 16 = 32 \\ \end{aligned}$

(-6/7)^2-2*16/7=-188/49 , because 1/a,1/b,1/c,1/d,1/e are the roots of equation 7x^5+6x^4+16x^3+7x^2-9x+1=0, then with Viet...