QUIZ MATH ~ IK Persamaan kuadrat (k + 2)x² - (2k - 1)x + k - 1 = 0 mempunyai akar-akar nyata dan sama. Jumlah kedua akar persamaan tersebut adalah ... . (UN 200

(k + 2)x² - (2k - 1)x + (k - 1) = 0

Syarat akar nyata dan sama
D = 0
b² - 4ac
(-2k + 1)² - 4(k + 2)(k - 1) = 0
(4k² - 4k + 1) - [(4k + 8)(k - 1)] = 0
4k² - 4k + 1 - (4k² + 4k - 8) = 0
-8k + 9 = 0
8k = 9
k = 9/8

Persamaan Kuadratnya menjadi...
(9/8 + 2)x² - (2.9/8 - 1)x + (9/8 - 1) = 0
(25/8)x² - (5/4)x + (1/8) = 0
25x² - 10x + 1 =

Berdasarkan Teorema Vietta
X1 + X2 = -b/a
X1 + X2 = -(-10)/25
X1 + X2 = 10/25
X1 + X2 = 2/5

a = k + 2
b = - (2k - 1) --> -2k + 1
c = k - 1

Akar nyata , dan sama (kembar) , maka :

D = 0
b² - 4ac = 0
(-2k + 1)² -4(k + 2)(k - 1) = 0
( 4k² - 4k + 1) - 4(k² + k - 2) = 0
4k² - 4k + 1 - 4k - 4k + 8 = 0
-8k + 9 = 0
9 = 8k
k = 9/8

Jumlah akar akar dr PK :
-b /a
= - ( - (2k - 1)) / ( k + 2)
= (2k - 1) / ( k + 2)
= ( ( 2 × 9/8) - 1) / ( (9/8) + 2)
= ( (9/4) - (4/4)) / ( (9/8) + 16/8)
= ( 5/4) / (25/8)
= 5/4 × 8/25
= 2/5