Question

Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial x²-2x-6​

Answer 1

Answer:

• 6x² + 2x - 1 = 0

Steps:

Let the zeros of the polynomial (x² - 2x - 6) be 'α' and 'β'

Hence sum of zeros would be:

→ α + β = -b/a

⇒ α + β = -(-2) / 1 = 2

→ αβ = c/a

⇒ αβ = (-6)/1 = -6

Now it is given that the zeros of the new polynomial is the reciprocals of the zeros of the above polynomial. Hence the zeros of the new polynomial are:

• 1/α (and) 1/β    ...(i)

The general form of a quadratic polynomial is written as:

• x² - (Sum of zeros) x + (Product of zeros)

According to (i), the sum of the new polynomial is:

$\implies \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{ (\alpha + \beta)}{\alpha \times \beta}$

We already know the values of (α+β) and (αβ). Hence we get:

$\implies \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{2}{-6} = \boxed{ \bf{\dfrac{-1}{3}}}$

Hence the sum of zeros of the new polynomial is ( -1/3 ).

Calculating the product of zeros we get:

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

Hence substituting the value of (αβ), we get:

$\implies \dfrac{1}{\alpha} \times \dfrac{1}{\beta} = \dfrac{1}{-6} = \boxed{\bf{\dfrac{-1}{6}}}$

Substituting these values in the general form we get:

$\implies x^2 - ( \dfrac{-1}{3} )x + (\dfrac{-1}{6})=0\\\\\\\implies x^2 + \dfrac{x}{3} - \dfrac{1}{6} = 0\\\\\\\text{Taking LCM we get:}\\\\\\\implies \dfrac{6x^2 + 2x - 1}{6} = 0\\\\\\\implies \boxed{ \bf{ 6x^2 + 2x - 1 = 0 }}$

Hence the required quadratic polynomial is: (6x² + 2x - 1 = 0)