persamaan kuadrat mempunyai akar akar

4x² - x - 5 = 0
(x + 1) (4x - 5)
x1 = -1 dan x2 = 5/4

x1² = (-1)² = 1
x2² = (5/4)² = 25/16

x1 + x2 = -b/a
1 + 25/16 = -b/a
41/16 = -b/a ~~~~> berarti b = - 41 dan a = 16

x1x2 = c/a
1 x 25/16 = c/a ~~~~> berarti c = 25

pers. baru = 16x² - 41x + 25 = 0

#maaf jika salah semoga membantu

Mapel : Matematika
Bab : Persamaan Kuadrat

Diketahui :
- 4x² - x - 5 = 0
- Akar 1 = x₁
- Akar 2 = x₂

Ditanya :
- Nilai x₁² + x₂² = ......?

Penyelesaian :
- Nilai x₁ dan x₂ :
4x² - x - 5 = 0
= 4x² + 4x - 5x - 5
= 4x(x + 1)- 5(x + 1)
= (4x - 5)(x + 1)

x₁ :
4x - 5 = p
4x = 5
x = 5/4

x₂ :
x + 1 = 0
x = - 1

- Persamaan Kuadrat Berakar x₁² dan x₂² :
a. Akar 1 :
x₁²
= (5/4)²
= 25/16

x = 25/16
16x = 25
16x - 25 = 0

b. Akar 2 :
x₂²
= (- 1)²
= 1

x = 1
x - 1 = 0

c. Persamaan Kuadrat :
(16x - 25)(x - 1)
= (16x × x) - (16x × 1) - (25 × x) + (25 × 1)
= 16x² - 16x - 25x + 25
= 16x² - 41x + 25

16x² - 41x + 25 = 0

Maaf Kalau Salah