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Verified answer
1) ditanyakan akar akar persamaana) -x² + 2x + 3 = 0
(-x + 3) (x + 1) = 0
x = 3 atau x = -1
b) x² + 8x + 12 = 0
(x + 2) (x + 6) = 0
x = -2 atau x = -6
2) dengan rumus ABC
a) 3x² - 6x - 3 = 0 (semua dibagi 3)
x² - 2x - 1 = 0
a = 1
b = -2
c = -1
x = (-b ± √b² - 4ac) / 2a
x = ( -(-2) ± √(-2)² - 4(1)(-1) / 2(1)
x = (2 ± √4 + 4) / 2
x = (2 ± √8) / 2
x = (2 ± 2√2) / 2
x = (1 ± √2)
maka akar persamaannya
x1 = 1 + √2 atau x2 = 1 - √2
b) 2x² - 13x + 15 = 0
a = 2
b = -13
c = 15
x = (-b ± √b² - 4ac) / 2a
x = ( -(-13) ± √(-13)² - 4(2)(15) ) / 2(2)
x = (13 ± √169 - 120) / 4
x = (13 ± √49) / 4
x = (13 ± 7) /4
maka akar persamaannya
x1 = (13 + 7) / 4 atau x2 = (13 - 7) / 4
x1 = (20) /4 atau x2 = 6 /4
x1 = 5 atau x2 = 3/2