a. 4xpangkat2 + 21x + 5 = 0
b. 6xpangkat2 – 11x + 3 = 0
Tentukan nilai 4a – 3b untuk a > b.
Diketahui a dan b merupakan akar akar persamaan kuadrat berikut.
(Persamaan 1)
4x²+21x+5=0
(4x+1)(x+5)=0
x= -1/4 =a
x= -5 =b
4(-1/4)+15=14
(Persamaan 2)
6x²-11x+3=0
(3x-1)(2x-3)=0
x=1/3 =b
x=3/2 =a
4a-3b= 3×2-3×(1/3)=6-1= 5
A. 4x^ + 21x + 5 = 0
( 4x + 1 ) ( x + 5 ) = 0
x, = -1/4 ====> a = -1/4
x,, = -5 ====> b = -5
4a – 3b = 4(-1/4) – 3(-5)
= -1 + 15 = 14
b. 6x^ – 11x + 3 = 0
( 2x – 3 ) ( 3x – 1 ) = 0
x, = 3/2 ===> a
x,, = 1/3 ===> b
4a – 3b = 4(3/2) – 3(1/3)
= 6 – 1 = 5
Semoga bisa membantu ya