Question

Jika persamaan kuadrat 3x^2 + x - 3 = 0 mempunyai akar akar á dan â maka persamaan kuadrat yang akar akarnya 2 + 1/a-1 dan 2 + 1/b-1 adalah

Answer 1

Penjelasan dengan langkah-langkah:

3x² +x - 3 = 0

a = 3

b = 1

c = -3

α + β = -b/a = - 1/3

α x β = c/a = -3/3 = -1

akat persamaan baru

2 + 1/(α + 1) dan

2 + 1/(β + 1)

x₁ + x₂

= 2+ 1/(1 + α ) + 2 + 1/(1 + β)

= 4 + 1/(1 + α ) + 1/(1 + β)

= 4+ [(1 + β) + (1 + α)]/(1 + α )(1 + β)

= 4 + [(2 + (a + b)/(αβ + α + β + 1)

= 4 + [2 + (-1/3)/(-1) + (-1/3) + 1]

= 4 + [ (2 - 1/3)/(-1/3)

= 4 + [(6/3 - 1/3)/(-1/3)

= 4 + (5/3)/(-1/3)

= 4 - 5

= - 1

x₁. x₂

= [2+ 1/(1 + α )] x [2 + 1/(1 + β)]

= 4 + 2/(1 + β) + 2/(1 + α) + 1/(1 +α)(1 + β)

= 4 + [2(1 + α) + 2(1 + β) + 1]/(1 +α)(1 + β)

= 4 + [(2 + 2α + 2 + 2β + 1)/(1 +α)(1 + β)

= 4 + [5 + 2(α + β)]/[(αβ + (α + β) + 1]

= 4 + [5 + 2(-1/3)]/(-1) + (-1/3) + 1]

= 4 + [(5 - 2/3)/(-1/3)

= 4 + [(15/3 - 2/3)/(-1/3)]

= 4 + (13/3)/(-1/3)

= 4 - 13

= - 9

persamaan kuadrat baru

x² - (x₁ + x₂)x + (x₁.x₂) = 0

x² - (-1)x + (-9) = 0

x² + x - 9 = 0