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1. Akar-akar persamaan kuadrat 3x2 + 2x - 5 = 0 adalah x1 dan x2. Hitunglah nilai dari 1/x1 + 1/x2.
Jawaban:
a = 3, b = 2, dan c = -5.
x1 + x2 = -b/a
⇒ x1 + x2 = -2/3
x1.x2 = c/a
⇒ x1 . x2 = -5/3
1/x1 + 1/x2 = (x1 + x2) / (x1.x2)
⇒ 1/x1 + 1/x2 = (-2/3) / (-5/3)
⇒ 1/x1 + 1/x2 = -2/3 . (-3/5)
⇒ 1/x1 + 1/x2 = 2/5
⇒ 1/x1 + 1/x2 = 0,4.
2. Jika x1 dan x2 adalah akar-akar dari persamaan kuadrat 2x2 - 6x - p = 0 dan x1 - x2 = 5, maka tentukanlah nilai p.
Jawaban:
a = 2, b = -6, dan c = -p.
x1 - x2 = (√D) / a
⇒ (x1 - x2) a = √D
⇒ (x1 - x2) a = √(b2 - 4.a.c)
⇒ 5(2) = √(36 - 4.2.(-p)
⇒ 10 = √(36 + 8p)
⇒ 100 = 36 + 8p
⇒ 8p = 64 ⇒ p = 8.
3. Jika x1 dan x2 merupakan akar dari persamaan 32x + 33-2x - 28 = 0, maka tentukanlah jumlah kedua akar tersebut.
Jawaban:
⇒ 32x + (33)/32x - 28 = 0
⇒ a + 27/a - 28 = 0
⇒ a2 - 27 - 28a = 0
⇒ a2 - 28a - 27 = 0
⇒ (a - 1)(a - 27) = 0
⇒ a = 1 atau a = 27
Untuk a = 1, maka : 32x = a
⇒ 32x =1
⇒ 32x = 30
⇒ 2x = 0
⇒ x1 = 0
Untuk a = 27, maka : 32x = a
⇒ 32x = 27
⇒ 32x = 33
⇒ 2x = 3
⇒ x2 = 3/2
Jadi x1 + x2 = 0 + 3/2 = 3/2.
4. Suatu persamaan kuadrat memiliki akar-akar x1 dan x2. Jika persamaan kuadrat tersebut adalah 2x2 - 3x - 5 = 0 , maka tentukanlah sebuah persamaan kuadrat baru yang akar-akarnya -1/x1 dan -1/x2.
Jawaban:
Dari persamaan kuadrat di soal diketahui a = 2, b = -3, dan c = -5.
x1 + x2 = -b/a
⇒ x1 + x2 = -(-3)/2
⇒ x1 + x2 = 3/2 x1.x2 = c/a
⇒ x1.x2 = -5/2
α = -1/ x1 dan β = -1/x2. α + β = (-1/x1) + (-1/x2)
⇒ α + β = (-1/x1) - (1/x2)
⇒ α + β = (-x2 - x1) / (x1.x2)
⇒ α + β = - (x1 + x2) / (x1.x2)
⇒ α + β = -(3/2) / (-5/2)
⇒ α + β = 3/5
α.β = -1/ x1 . (-1/x2)
⇒ α.β = 1/(x1.x2)
⇒ α.β = 1/ (-5/2)
⇒ α.β = -2/5
Jadi persamaan kuadrat yang akarnya -1/ x1 dan -1/x2 adalah:
x2 - (α + β)x + α.β = 0
⇒ x2 - 3/5x + (-2/5) = 0
⇒ x2 - 3/5x - 2/5 = 0
⇒ 5x2 - 3x - 2 = 0.
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