Tentukan akar Persamaan berikela 3x² – 12 = 06 X² – 7x+6 – 0​

Jawaban:

Akar-akar persamaan berikut :

a) 3x – 12 = 0

=> 3x = 12

=> x = 12/3

=> x = 4

HP = {4}

b) x² + 7x + 6 = 0

=> x² + x + 6x + 6 = 0

=> x(x + 1) + 6(x + 1) = 0

=> (x + 6)(x + 1) = 0

=> (x + 6) = 0 atau (x + 1) = 0

=> x = -6 atau x = -1

HP = {-6, -1}

c) -3x² – 5x + 2 = 0

=> -3x² – 6x + x + 2 = 0

=> -3x(x + 2) + 1(x + 2) = 0

=> (x + 2)(-3x + 1) =

=> (x + 2) = 0 atau (-3x + 1) = 0

=> x = -2 atau x = 1/3

HP = {-2, 1/3}

2. Nyatakan persamaan 3(x² + 1) + x(x – 3) dalam bentuk umum persamaan kuadrat

Bentuk umum persamaan kuadrat :

ax² + bx + c = 0 dengan a ≠ 0

3(x² + 1) + x(x – 3) = 0

3x² + 3 + x² – 3x = 0

4x² – 3x + 3 =