#janganjawabasalasalan
Jawab:
Terlampir pada penjelasan.
Penjelasan dengan langkah-langkah:
Materi : Matematika - Persamaan Kuadrat
Rumus abc = [tex]x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}[/tex]
Soal 1
a = 1
b = 7
c = 12
[tex]x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-7+/-\sqrt{7^{2}-4(1)(12)} }{2*1}\\\\=\frac{-7+/-\sqrt{49-48} }{2} \\=\frac{-7+/-1}{2} \\\frac{-7+(-1)}{2} = \frac{-8}{2} = -4\\\frac{-7-(-1)}{2} = \frac{-7+1}{2} = \frac{-6}{2} = -3[/tex]
Soal 2
a = 1
b = 10
c = 16
[tex]x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-10+/-\sqrt{10^{2}-4(1)(16)} }{2*1}\\\\=\frac{-10+/-\sqrt{100-64} }{2} \\=\frac{-10+/-\sqrt{36} }{2} \\\frac{-10+6}{2} = \frac{-4}{2} = -2\\\frac{-10-(6)}{2} = \frac{-16}{2} = -8[/tex]
Soal 3
a = 1
b = -8
c = 12
[tex]x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(-8)+/-\sqrt{(-8)^{2}-4(1)(12)} }{2*1}\\\\=\frac{8+/-\sqrt{64-48} }{2} \\=\frac{8+/-\sqrt{16} }{2} \\\frac{8+4}{2} = 6\\\\\frac{8-(4)}{2} = 1[/tex]
Soal 4
a = 1
b = -9
c = 18
[tex]x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(-9)+/-\sqrt{(-9)^{2}-4(1)(18)} }{2*1}\\\\=\frac{9+/-\sqrt{81-72} }{2} \\=\frac{9+/-\sqrt{9} }{2} \\\frac{9+3}{2} = 6\\\\\frac{9-(3)}{2} = 3[/tex]
Soal 5
a = 2
b = 3
c = -9
[tex]x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(3)+/-\sqrt{(3)^{2}-4(2)(-9)} }{2*2}\\\\=\frac{-3+/-\sqrt{9+72} }{4} \\=\frac{-3+/-\sqrt{81} }{4} \\\frac{-3+9}{4} = \frac{3}{2} \\\\\frac{-3-(9)}{4} = -3[/tex]
Soal 6
a = 3
b = 13
c = -10
[tex]x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(13)+/-\sqrt{(13)^{2}-4(3)(-10)} }{2*3}\\\\=\frac{-13+/-\sqrt{169+120} }{6} \\=\frac{-13+/-\sqrt{289} }{6} \\\frac{-13+17}{6} = \frac{2}{3}\\\\\frac{-13-(17)}{6} = -5[/tex]
Soal 7
a = 4
b = -8
c = -5
[tex]x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(-8)+/-\sqrt{(-8)^{2}-4(4)(-5)} }{2*4}\\\\=\frac{8+/-\sqrt{64+80} }{8} \\=\frac{8+/-\sqrt{144} }{8} \\\frac{8+12}{8} = \frac{5}{2}\\\\\frac{(8-12)}{8} = -\frac{1}{2}[/tex]
Soal 8
a = 6
b = -7
c = -3
[tex]x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(-7)+/-\sqrt{(-7)^{2}-4(6)(-3)} }{2*6}\\\\=\frac{7+/-\sqrt{49+72} }{12} \\=\frac{7+/-\sqrt{121} }{12} \\\frac{7+11}{12} = \frac{3}{2}\\\\\frac{(7-11)}{12} = -\frac{1}{3}[/tex]
Soal 9
a = 4
b = 20
c = 25
[tex]x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(20)+/-\sqrt{(20)^{2}-4(4)(25)} }{2*4}\\\\=\frac{-20+/-\sqrt{400-400} }{8} \\=\frac{-20+/-\sqrt{0} }{8} \\\frac{-20}{8} = -\frac{5}{2}[/tex]
Soal 10
a = 9
b = -30
c = 25
[tex]x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(-30)+/-\sqrt{(-30)^{2}-4(9)(25)} }{2*9}\\\\=\frac{30+/-\sqrt{900-900} }{18} \\=\frac{30+/-\sqrt{0} }{18} \\\frac{30}{18} = \frac{15}{9}=\frac{5}{3}[/tex]
Rujukan : Matematika, Kelas VIII SMP KTSP 2006 (2008)
Selamat belajar.
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