Akar-akar Persamaan Linear

Persamaan Linear Pangkat 1 (Satu)

• Persamaan: y = f(x) = ax + b
  x = variabel bebas
  y = variabel tak bebas (terikat)
  a = koefisien/gradien/kemiringan
  b = konstanta
• Gambar:
  

Persamaan Linear Pangkat 2 (Dua)

• Persamaan: y = f(x) = ax^2 + bx + c
• Cari akar pakai rumus: x1,2 = [-b ± akar (b2 – 4ac)] : 2a
• Gambar:
  

Contoh Soal
Tentukan akar-akar persamaan linear sebagai berikut:
1. x^2 – 5x – 24 = 0
2. y = 3x^2 – 6x – 26
3. y = -2x^2 + 8x +30
4. y = 15 – 23x + 6x^2

jawab:

1. x^2 – 5x – 24 = 0
   maka a = 1; b = -5; c = -24
   x1,2 = [-b ± akar (b^2 – 4ac)] : 2a
   = [-(-5) ± akar ((-5)^2 – 4.1.(-24))] : 2.1
   = [5 ± akar (25 – (-96))] : 2
   = [5 ± akar (25 + 96)] : 2
   = [5 ± akar 121] : 2
   = [5 ± 11] : 2
   maka : x1 = [5 + 11] : 2 = 16 : 2 = 8
   : x2 = [5 – 11] : 2 = -6 : 2 = -3q
   
   
2. y = 3x^2 – 6x – 26
   x1,2 = [-(-6) ± akar ((-6)^2 – 4.3.(-26))] : 2.3
   = [6 ± akar (36 + 312)] : 6
   = [6 ± akar 348] : 6
   = [6 ± 18,65] : 6
   maka : x1 = [6 + 18,65] : 6 = 4,11
   x2 = [6 – 18,65] : 6 = -2,11
   
   
3. y = -2x^2 + 8x +30
   x1,2 = [-(8) ± akar (8^2 – 4.(-2).30)] : 2.(-2)
   = [-8 ± akar (64 + 240)] : -4
   = [-8 ± akar 304] : -4
   = [-8 ± 17,44] : -4
   maka : x1 = [-8 + 17,44] : -4 = -2,36
   x2 = [-8 – 17,44] : -4 = 6,36
   
   
4. y = 15 – 23x + 6x^2
   x1,2 = [-(-23) ± akar ((-23)^2 – 4.6.15)] : 2.6
   = [23 ± akar (529 – 360) : 12
   = [23 ± akar 169] : 12
   = [23 ± 13] : 12
   maka : x1 = [23 + 13] : 12 = 3
   x2 = [23 – 13] : 12 = 0,83

Tugas Mata Kuliah Metode Numerik
Rabu, 02 Desember 2009