jka persamaan persamaan kuadrat x^2-ax=4=0 danx^2=6x=b=0 akar kembar dengan syarat ab>0, maka nilai 3a-4b=

Ax^2 + Bx + C = 0 mempunyai akar kembar jika D = 0
=> B^2 - 4AC = 0
1) x^2 - ax + 4 = 0
D = (-a)^2 - 4(1)(4) = 0
=> a^2 - 16 = 0
=> a^2 = 16
=> a = ±4
=> a = 4 atau a = -4

2) x^2 + 6x + b = 0
D = 6^2 - 4(1)(b) = 0
=> 36 - 4b = 0
=> -4b = -36
=> b = 9

Karena ab > 0 maka a = 4 dan b = 9
3a - 4b = 3(4) - 4(9) = 12 - 36 = -24

x² - ax + 4 = 0

Syarat akar kembar
D = 0
b² - 4ac = 0
(-a)² - 4(1)(4) = 0
a² - 16 = 0
a = 4
a = -4, TM karena ab > 0

x² + 6x + b = 0

Syarat akar kembar
D = 0
b² - 4ac = 0
6² - 4(1)(b) = 0
36 - 4b = 0
4b = 36
B = 9

Maka...
3a - 4b
= 3(4) - 4(9)
= 12 - 36
= -24