tentukan penyelesaian dari persamaan-persamaan berikut dengan menggunakan rumus abc!#janganjawabasal.... Question from @ijklmnoabcdefgh12

ijklmnoabcdefgh12 @ijklmnoabcdefgh12

October 2022 1 5 Report

tentukan penyelesaian dari persamaan-persamaan berikut dengan menggunakan rumus abc!

#janganjawabasalasalan

Griccie

Jawab:

Terlampir pada penjelasan.

Penjelasan dengan langkah-langkah:

Materi : Matematika - Persamaan Kuadrat

Rumus abc = $x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}$

Soal 1

a = 1

b = 7

c = 12

$x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-7+/-\sqrt{7^{2}-4(1)(12)} }{2*1}\\\\=\frac{-7+/-\sqrt{49-48} }{2} \\=\frac{-7+/-1}{2} \\\frac{-7+(-1)}{2} = \frac{-8}{2} = -4\\\frac{-7-(-1)}{2} = \frac{-7+1}{2} = \frac{-6}{2} = -3$

Soal 2

a = 1

b = 10

c = 16

$x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-10+/-\sqrt{10^{2}-4(1)(16)} }{2*1}\\\\=\frac{-10+/-\sqrt{100-64} }{2} \\=\frac{-10+/-\sqrt{36} }{2} \\\frac{-10+6}{2} = \frac{-4}{2} = -2\\\frac{-10-(6)}{2} = \frac{-16}{2} = -8$

Soal 3

a = 1

b = -8

c = 12

$x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(-8)+/-\sqrt{(-8)^{2}-4(1)(12)} }{2*1}\\\\=\frac{8+/-\sqrt{64-48} }{2} \\=\frac{8+/-\sqrt{16} }{2} \\\frac{8+4}{2} = 6\\\\\frac{8-(4)}{2} = 1$

Soal 4

a = 1

b = -9

c = 18

$x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(-9)+/-\sqrt{(-9)^{2}-4(1)(18)} }{2*1}\\\\=\frac{9+/-\sqrt{81-72} }{2} \\=\frac{9+/-\sqrt{9} }{2} \\\frac{9+3}{2} = 6\\\\\frac{9-(3)}{2} = 3$

Soal 5

a = 2

b = 3

c = -9

$x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(3)+/-\sqrt{(3)^{2}-4(2)(-9)} }{2*2}\\\\=\frac{-3+/-\sqrt{9+72} }{4} \\=\frac{-3+/-\sqrt{81} }{4} \\\frac{-3+9}{4} = \frac{3}{2} \\\\\frac{-3-(9)}{4} = -3$

Soal 6

a = 3

b = 13

c = -10

$x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(13)+/-\sqrt{(13)^{2}-4(3)(-10)} }{2*3}\\\\=\frac{-13+/-\sqrt{169+120} }{6} \\=\frac{-13+/-\sqrt{289} }{6} \\\frac{-13+17}{6} = \frac{2}{3}\\\\\frac{-13-(17)}{6} = -5$

Soal 7

a = 4

b = -8

c = -5

$x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(-8)+/-\sqrt{(-8)^{2}-4(4)(-5)} }{2*4}\\\\=\frac{8+/-\sqrt{64+80} }{8} \\=\frac{8+/-\sqrt{144} }{8} \\\frac{8+12}{8} = \frac{5}{2}\\\\\frac{(8-12)}{8} = -\frac{1}{2}$

Soal 8

a = 6

b = -7

c = -3

$x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(-7)+/-\sqrt{(-7)^{2}-4(6)(-3)} }{2*6}\\\\=\frac{7+/-\sqrt{49+72} }{12} \\=\frac{7+/-\sqrt{121} }{12} \\\frac{7+11}{12} = \frac{3}{2}\\\\\frac{(7-11)}{12} = -\frac{1}{3}$

Soal 9

a = 4

b = 20

c = 25

$x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(20)+/-\sqrt{(20)^{2}-4(4)(25)} }{2*4}\\\\=\frac{-20+/-\sqrt{400-400} }{8} \\=\frac{-20+/-\sqrt{0} }{8} \\\frac{-20}{8} = -\frac{5}{2}$

Soal 10

a = 9

b = -30

c = 25

$x_1,_2=\frac{-b+/-\sqrt{b^{2}-4ac} }{2a}\\= \frac{-(-30)+/-\sqrt{(-30)^{2}-4(9)(25)} }{2*9}\\\\=\frac{30+/-\sqrt{900-900} }{18} \\=\frac{30+/-\sqrt{0} }{18} \\\frac{30}{18} = \frac{15}{9}=\frac{5}{3}$