persamaan kuadrat 2x²+x-4=0. akar akarnya x1 dan x2. persamaan kuadrat baru yang akar akarnya (x1-4) dan (x2-4) adalah

Kelas 10 Matematika
Bab Persamaan Kuadrat

x - 4 = a
x = a + 4

2x² + x - 4 = 0
2 (a + 4)² + (a + 4) - 4 = 0
2 (a² + 8a + 16) + a + 4 - 4 = 0
2a² + 16a + 32 + a = 0
2a² + 17a + 32 = 0
2x² + 17x + 32 = 0

2x^2 + x - 4 =0
a = 2
b = 1
C = -4

X1 + x2 = -b/a = -1/2
X1.X2 = c/a = -4/2 = -2

(X1-4) + (X2-4) = (X1+X2)-8 = -1/2 -8 = -17/2

(X1-4) . (X2-4) = x1.x2-4x1-4x2+16= (x1.x2) - 4(x1+x2) + 16 = -2-4(-1/2)+16 = 16

X^2 - {(X1-4) + (X2-4)} + {(X1-4) . (X2-4)} = 0
x2 + 17/2x + 16 =0 (seluruhnya dibagi 2)
1/2x + 17x + 8 =0