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Divina A
12 September 2022 07:08
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DA
Divina A
12 September 2022 07:08
Pertanyaan
Tentukan akar akar persamaan kuadrat 2x²-5x-12=0 pakai rumus ABC !
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NR
Nazhifah R
12 September 2022 08:02
2x²-5x-12=0 a = 2, b = -5, c = -12 / = per Rumus abc x1,2 = -b±√b²-4ac / 2a x1,2 = -(-5)±√(-5)²-4(2)(-12) / 2(2) x1,2 = 5±√25+96 / 4 x1,2 = 5±√121 / 4 x1,2 = 5±11 / 4 x1 = 5+11 / 4 x1 = 16 / 4 x1 = 4 Atau x2 = 5-11 / 4 x2 = -6 / 4 x2 = -3 / 2
· 4.0 (4)
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DI
Dina I
12 September 2022 08:03
<h1>2x²-5x-12=0</h1><p>a = 2 b = -5 c = -12</p><p> </p><p>= { (-b) ±√(b<sup>2</sup>-4ac)} / 2a</p><p>= {(-5) ± √((-5<sup>2</sup>)-4×2×(-12)} / 2×2</p><p>= {5 ± √(25)-(-96)} /4</p><p>= (5 ± √121) / 4</p><p>= (5±11)/4</p><p> </p><p>Makan Nilai x yaitu:</p><p>X<sub>1 </sub>= (5+11)/4 = 16/4 = 4</p><p>X<sub>2 </sub>= (5-11)/4 = -6/4 = -3/2 = -1,5</p><p>Jadi HP : {4, -1,5}</p><p> </p>
2x²-5x-12=0
a = 2 b = -5 c = -12
= { (-b) ±√(b2-4ac)} / 2a
= {(-5) ± √((-52)-4×2×(-12)} / 2×2
= {5 ± √(25)-(-96)} /4
= (5 ± √121) / 4
= (5±11)/4
Makan Nilai x yaitu:
X1 = (5+11)/4 = 16/4 = 4
X2 = (5-11)/4 = -6/4 = -3/2 = -1,5
Jadi HP : {4, -1,5}
· 5.0 (2)
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