Question

if a & ß
are zeros of polynomial
$6{y}^{2}$-7y+2, find
quadratic polynomial whose
zeros are 1/
$\alpha$
and 1/
$\beta$
​

Answer 1

Answer:

Step-by-step explanation

Roots of equation 6y²-7y+2=

6y²-7y+2=0

6y²-4y-3y+2=0

2y(3y-2)-1(3y-2)=0

(2y-1)(3y-2)=0

y=1/2 or y=2/3

Let α=1/2 and β=2/3

Therefore,1/α=2 and 1/β=3/2

Standard equation= x²-(1/α+1/β)x+1/αβ (in this question)

                              = x²-(2+3/2)x+1/2.3/2

                              =x²-7/2x+3/4

                              =4x²-14x+3