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

jawab

(p-1)x² - 6x + (r  + 4) = 0
(2q - 1)x² + px - (q+1)= 0
kedua akar salingf berkebalikan

(r+4)x² - 6x +(p-1) = (2q-1)x² + px - (q+1)

r + 4 = 2q - 1 ...(i)
-6 = p...............(ii)
p-1 = - (q -1)....(iii)
..
(ii)  -6 = p
p = - 6

(iii)  p-1 = - (q  - 1)
-6 - 1 = - q + 1
- 7 = - q + 1
q = 1 + 7
q = 8

(i) r + 4 = 2q - 1
r = 2q -1- 4
r =2(8) - 5
r = 11

p + q + r = (-6)+(8) +(11)= 13