​ If α and β are the zeros of the polynomial f(x) = x^2 + px + q, then a polynomial having 1/α and 1/β and is its zeros is

Question

If α and β are the zeros of the polynomial f(x) = x² + px + q, then a polynomial having \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) is its zeros is

A. x² + qx + p

B. x² – px + q

C. qx² + px + 1

D. px² + qx + 1

Answer 1

Given:

If α and β are the zeros of the polynomial

f(x) = x² + px + q,

To find:

A polynomial having \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) is its zeros is

Solution:

f(x) = x² + px + q

We know,

Since α, β are zeroes of given polynomial,

⇒ α + β = – p and αβ = q

Let S and P denote respectively the sum and product of zeroes of the required polynomial,

So,

Put the values of α + β and αβ in (1) and (2) to get,

⇒ S = \(\frac{-p}{q}\)

And

P = \(\frac{1}{q}\)

We know equation having 2 zeroes is of form,k (x² - (sum of zeroes) x + product of zeroes)

For a polynomial having \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\)is its zeros the equation becomes,

x² + p/q x + 1/q = 0

So here we get,

g(x) = qx² + px + 1