let f(x) = x2-6x+k
now ,3α + 2β = 20 ……………….(i)
α + 2α + 2β = 20
α +2(α + β) = 20
α + 2(-b/a) = 20
α + 2 { -(-6)/1} = 20
α + 2 x 6 = 20 => α + 12 = 20 => α = 20-12
α = 8
substituting this value of α in (i)
3 α +2 β = 20 => 3 x 8 + 2 β = 20
2 β = -4 => β = -2
α β = c/a => 8 x -2 = k/1 => k/1 = -16
So , k= -16