Question

Natural numbers p, q, r, s are such that if α, β are the roots of x2 – px + q = 0 then (α+1/β) and (β+1/α) are the roots of x2 – rx + s = 0. What is the possible value of s? Answer:

Answer 1

$\underline{\textbf{Given:}}$

$\mathrm{\alpha\;\&\;\beta\;are\;roots\;of\;x^2-p\,x+q=0\;and}$

$\mathrm{\dfrac{\alpha+1}{\beta}\;\&\;\dfrac{\beta+1}{\alpha}\;are\;roots\;of\;x^2-r\,x+s=0}$

$\underline{\textbf{To find:}}$

$\text{The value of 's'}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Concept used:}}$

$\mathrm{If\;\alpha\;\&\;\beta\;are\;roots\;ax^2+bx=c=0,\;then}$

$\mathrm{\alpha+\beta=\dfrac{-b}{a}\;\;\&\;\;\alpha\beta=\dfrac{c}{a}}$

$\mathrm{Since\;\alpha\;\&\;\beta\;are\;roots\;of\;x^2-p\,x+q=0,}$

$\mathrm{\alpha+\beta=p\;\;\&\;\;\alpha\beta=q}$

$\mathrm{Since\;\dfrac{\alpha+1}{\beta}\;\&\;\dfrac{\beta+1}{\alpha}\;are\;roots\;of\;x^2-r\,x+s=0}$

$\mathrm{Product\;of\;roots=s}$

$\mathrm{\dfrac{\alpha+1}{\beta}{\times}\dfrac{\beta+1}{\alpha}=s}$

$\mathrm{\dfrac{\alpha\beta+\alpha+\beta+1}{\alpha\beta}=s}$

$\mathrm{\dfrac{q+p+1}{q}=s}$

$\implies\boxed{\mathrm{\bf\,s=\dfrac{p+q+1}{q}}}$

$\underline{\textbf{Find more:}}$

if α & β are the zeros of the quadratic polynomial p(x)= 3X²-6x+4,find the value of α\β+β+α+2(1/α+1/β)+3αβ​

https://brainly.in/question/18209208