Question

if α and β are zeroes of polynomial f(x)=x^2 +px +q then polynomial having 1/α and 1/β as its zero is ​

Answer 1

Answer:

α+ β = -p αβ = q

(1/α + 1/β) = (α + β) / αβ  = - p / q

1/αβ  = 1 / q.

If 1/α, 1/β are zeros of the quadratic polynomial then the equation is

x2 -(1 / α + 1 / β)x + 1 / αβ = 0 then

x2 -(-p / q)x + 1 / q = 0

qx2 + px + 1  = 0

Answer 2

Given:

α and β are zeroes of polynomial f(x)=x^2 +px +q.

To find:

The polynomial that has 1/α and 1/β as its zeroes.

Solution:

First of all, we need to know that if a polynomial ax^2 +bx +c has α and β as its zeroes, then

$α + β = \frac{ - b}{a}$

$α \times β = \frac{c}{a}$

where a, b are coefficients of x^2 and x respectively.

So, in the polynomial f(x)=x^2 +px +q, the sum of zeroes and product of zeroes are

$α + β = \frac{ - p}{1} = - p \: (i)$

$α \times β = \frac{ q}{1} = q \: \: (ii)$

Now,

we will find out the polynomial p(x) that has 1/α and 1/β as its zeroes.

So,

the sum of zeroes of p(x)

$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β}$

$= \frac{ - p}{q} \: \: \: from \: (i) \: and \: (ii)$

and

product of zeroes of p(x)

$\frac{1}{α} \times \frac{1}{β} = \frac{1}{α β}$

$= \frac{1}{q} \: \: \: \: from \: (ii)$

Also, a quadratic polynomial p(x) is equal to

${x}^{2} - (sum \: of \: zeroes)x + (product \: of \: zeroes)$

So, the required polynomial p(x) having 1/α and 1/β as its zeroes is:

${x}^{2} - ( \frac{ - p}{q} )x + \: \frac{1}{q}$

$q {x}^{2} + px + 1$

Hence, the polynomial having 1/α and 1/β as its zeroes is qx2 +px +1.