Question

If alpha and beta are the zeros of the polynomial 6y^2 - 7y +2 , find a quadratic polynomial whose zeros are 1 upon alpha + 1 upon beta

Answer 1

Hi Mate !!

Solution :

P ( x ) = 6y² - 7y + 2

Here,

a = Coefficient of y² = 6

b = Coefficient of y = -7

and, c = Constant term = 2

Therefore,

sum of zeroes = -b/a

alpha + beta = -(-7) / 6

alpha + beta = 7/6 ---------(1)

and, product of zeroes = c/a

alpha × beta = 2 / 6 --------(2)

Sum of zeroes of new quadratic polynomial = 1/alpha + 1/beta = beta + alpha / alpha beta = 7/6 / 2/6 [ From equation (1) and (2) ]

=> 7/6 × 6/2 = 7/2.

And , product of zeroes of new quadratic polynomial = 1/alpha × 1/beta = 1 / alpha beta = 1/ 1 / 2/6 = 1/1 × 6/2 = 6/2 = 3.

Therefore , Required quadratic polynomial = x² - ( sum of zeroes )x + product of zeroes.

=> x² - 7/2 x + 3

=> x² - 7x/2 + 3

=> 2x² - 7x + 6 [ Answer]

Hope it helps you ♥

By Rishi403.

BeBrainly

Answer 2

Hii ☺ !!

Here is your answer :

P ( x ) = 6y² - 7y + 2

Sum of zeroes ( S ) = - Coefficient of x / Coefficient of x².

Sum of zeroes ( S ) = 7 / 6

And,

Product of zeroes ( P ) = Constant term / Coefficient of x² .

Product of zeroes ( P ) = 2 / 6

Therefore,

Sum of zeroes of new quadratic polynomial (S1) = 1 / Alpha + 1/ß = ẞ + Alpha / Alpha ẞ = 7 / 6 / 2 / 6 = 7/2.

And,

Product of zeroes ( P1) = 1 / Alpha × 1/ ẞ= 1 / Alpha ẞ = 1 / 7 / 2 = 1 / 1 / 2/6 = 6/2 = 3.

Required quadratic polynomial = X² - ( Sum of zeroes ) X + Product of zeroes.

=> X² - ( S1 ) X + ( P1 )

=> X² - 7x/2 +3

=> 2x² - 7x + 6 [ Answer ]


☺ Hope it will help you ☺