Question

$obtain \: a \: quadratic \: polynomial \\ whose \: zeroes \: are \: \alpha + 2 \beta \: and \: \\ 2 \alpha + \beta \: where \: \alpha \: and \: \beta \: are \: the \: \\ zeroes \: of \: polynomial \ 3x {}^{2}- x -1.$

Answer 1

Answer:

9x² - 9x - 1 = 0

Step-by-step explanation:

Given Quadratic equation is 3x² - x - 1.

Here, a = 3, b = -1, c = -1.

Given that α,β are zeroes of polynomial.

(i) Sum of zeroes:

α + β = -b/a

        = 1/3

(ii) Product of zeroes:

αβ = c/a

     = -1/3.

Given Zeroes are α + 2β and 2α + β.

(i) Sum of zeroes:

2α + β + 2α + β

3α + 3β

3(α + β)

3(1/3)

1.

(ii) Product of zeroes:

= (α + 2β)(2α + β)

= 2α(α + 2β) + β(α + 2β)

= 2(α² + β²) + 5αβ

= 2(α + β)² + αβ

= 2(1/3)² + (-1/3)

= (2/9) - 1/3

= -1/9

∴ Required Quadratic Polynomial = x² - (Sum of zeroes)x + (product of zeroes) = 0

= x² - (1)x + (-1/9) = 0

⇒ 9x² - 9x - 1 = 0

Hope it helps!