Question

Find a quadratic polynomial whose zeroes are
reciprocals of the zeroes of the polynomial
f(x)=ax2 +bx +c
​

Answer 1

Answer :-

◾As we have to find out the Polynomial whose zeros Rare reciprocal of the Polynomial f(x) ax² + bx + c

Well first of all we should know that how to get the Sum and Product relation of roots

Let a quadratic equation be

$\sf{n(x-\alpha)(x-\beta)}$

◾Where $\sf{\alpha \:and \:\beta}$are the roots of the Polynomial

◾On simplification we get

$\sf{ n( x^2 -(\alpha+\beta)x + (\alpha\beta)}$

$\sf{ nx^2 -(\alpha+\beta)nx + (\alpha\beta)n}$

◾So we can say that

▪️Sum of roots $= \dfrac{-(\alpha+\beta)}{n}$

▪️Product of roots $= \dfrac{\alpha\beta}{n}$

◾ Now by substituting values of Coefficient

▪️Sum of roots $= -\dfrac{(b)}{a}$

▪️Product of roots $= \dfrac{c}{a}$

◾ Now the reciprocal of roots

▪️Sum of roots $= \dfrac{1}{\alpha} + \dfrac{1}{\beta}$

$= \dfrac{\alpha + beta }{\alpha \beta}$

$= \dfrac{ -\dfrac{(b)}{a}}{ \dfrac{(c)}{a}}$

$= \dfrac{-b}{c}$

▪️ Product of roots $= \dfrac{1}{\alpha} \times \dfrac{1}{\beta}$

$= \dfrac{1}{\alpha\beta}$

$= \dfrac{1}{c}$

◾Now Polynomial

$= k( x^2 + \dfrac{b}{c}x + \dfrac{1}{c})$