Question

If alfa and Beta zeroes of the quadratic polynomial p(x) = 6x² + x-1, then find the value of
alfa/Beta+beta/alfa+ 2(1/alfa+1/beta)+3alfa*beta​

Answer 1

p(x) = 6x^2 + x - 1

= 6x^2 + 3x - 2x - 1

= (6x^2 + 3x) - (2x + 1)

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

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

3x - 1 = 0

3x = 1

Alpha = x = 1/3

2x + 1 = 0

2x = -1

Beta = x = -1/2

(Alpha/Beta)+(Beta/Alpha)+2((1/Alpha)+(1/Beta))+3(Alpha)(Beta)

= ((1/3)/(-1/2))+((-1/2)/(1/3))+2((1/(1/3))+(1/(-1/3))+3(1/3)(-1/2)

= (-2/3)+(-3/2)+2((3/1)+(-3/1))+(1)(-1/2)

= ((-4-9)/6)+2((3)+(-3))+(-1/2)

= (-11/6)+2(0)+(-1/2)

= (-11/2)+0-(1/2)

= (-11/2)+(-1/2)

= (-11-1)/2

= -12/2

= -6

Required Answer is -6

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= -6