Persamaan kuadrat x^-bx+b-1=0 mempunyai dua akar real x1 dan x2. Jika 2x1+x2=65, maka nilai b adalah..

jawab

x² - bx + (b-1) = 0

x1+ x2 = b ......(i)
x1 x2 = b - 1....(ii)
2x1 + x2 = 65..(iii)

2x1 + x2 = 65
x1 + x 2 = b......(-)
x1 = 65 - b 

x1 + x2 = b
x2= b - x1
x2 = b- (65 - b)
x2 = 2b- 65

x1 x2 = b -1
(65-b)(2b -65) = b - 1
130b - 4225 - 2b² + 65b = b -1
-2b² + 194 b - 4225 = b - 1
- 2b² + 194b - 4224 = 0
b² - 97b + 2112  = 0
(b - 33)(b- 64) = 0
b = 33  atau b  = 64