Jawab:Akar-akar dari 3x² + 21x + 6 = 0 adalah:x₁ = (–7 + √41) / 2 , x₂ = (–7 – √41) / 2
Penjelasan dengan langkah-langkah:
Cara 1: Pemfaktoran
3x² + 21x + 6 = 0(kedua ruas dibagi 3)⇒ x² + 7x + 2 = 0
Melengkapkan persamaan kuadrat sempurna.⇒ x² + 7x = –2⇒ x² + 7x + 49/4 = –2 + 49/4⇒ [x + 7/2]² = –8/4 + 49/4⇒ [x + 7/2]² = 41/4⇒ x + 7/2 = ± √(41/4)⇒ x = –7/2 ± (√41)/2⇒ x₁ = (–7 + √41) / 2 , x₂ = (–7 – √41) / 2
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Cara 2: Rumus ABC
3x² + 21x + 6 = 0
(kedua ruas dibagi 3)⇒ x² + 7x + 2 = 0 ⇒ a = 1, b = 7, c = 2
Rumus ABCx = [–b ± √(b² – 4ac)] / 2a⇒ x = [–7 ± √(7² – 4·1·2)] / 2·1⇒ x = [–7 ± √(49 – 8)] / 2⇒ x = [–7 ± √41] / 2⇒ x₁ = (–7 + √41) / 2 , x₂ = (–7 – √41) / 2
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Jawab:
Akar-akar dari 3x² + 21x + 6 = 0 adalah:
x₁ = (–7 + √41) / 2 , x₂ = (–7 – √41) / 2
Penjelasan dengan langkah-langkah:
Cara 1: Pemfaktoran
3x² + 21x + 6 = 0
(kedua ruas dibagi 3)
⇒ x² + 7x + 2 = 0
Melengkapkan persamaan kuadrat sempurna.
⇒ x² + 7x = –2
⇒ x² + 7x + 49/4 = –2 + 49/4
⇒ [x + 7/2]² = –8/4 + 49/4
⇒ [x + 7/2]² = 41/4
⇒ x + 7/2 = ± √(41/4)
⇒ x = –7/2 ± (√41)/2
⇒ x₁ = (–7 + √41) / 2 , x₂ = (–7 – √41) / 2
________________
Cara 2: Rumus ABC
3x² + 21x + 6 = 0
(kedua ruas dibagi 3)
⇒ x² + 7x + 2 = 0 ⇒ a = 1, b = 7, c = 2
Rumus ABC
x = [–b ± √(b² – 4ac)] / 2a
⇒ x = [–7 ± √(7² – 4·1·2)] / 2·1
⇒ x = [–7 ± √(49 – 8)] / 2
⇒ x = [–7 ± √41] / 2
⇒ x₁ = (–7 + √41) / 2 , x₂ = (–7 – √41) / 2