Persamaan kuadrat (p-1)x^2 - 6x + (r+4) = 0 dan (2q -1 ) x^2 +px - (q+1) = 0 akar2nya saling berkebalikan, maka p+q+r = . ...

(p - 1)x^2 - 6x + (r + 4) = 0

persamaan kuadrat yang akar2 kebalikan dari persamaan di atas adalah
x^2 - (1/x1 + 1/x2)x + 1/(x1x2) = 0
x^2 - (x1 + x2)/(x1x2) x + 1/(c/a) = 0
x^2 - (-b/a)/(c/a) x + a/c = 0
x^2 + (b/c) x + (a/c) = 0
cx^2 + bx + a = 0
(r + 4)x^2 + (-6)x + (p - 1) = 0 => (2q - 1)x^2 + px - (q + 1) = 0
1) -6 = p
2) (p - 1) = -(q + 1)
=> -6 - 1 = -q - 1 => q = 6
3) r + 4 = 2q - 1
=> r = 2q - 5 = 2(6) - 5 = 7
jadi p + q + r = -6 + 6 + 7 = 7