Question

if alpha beta zeros of the polynomial 6x^2+x-1,then find the value of alpha/beta+beta/alpha (1/alpha+1/beta)+3 alpha beta ​

Answer 1

$6 {x}^2 + 3x - 2x - 1$

3x(2x+1)-1(2x+1)

2x+1(3x-1)=0

2x=-1,3x=1

x=-1/3.1/3

a=-1/3

b=1/3

a/b+b/a=a2+b2/ab=

1/9+1/9/-1/9

2/9*-9/1

=-2

and

1/a+1/b+3ab

=-1/3+1/3+3*1/3*-1/3

0+-1/9

=-1/9