Question

Tentukan akar persamaan kuadrat dengan rumus abc dari persamaan
a. x²-2x=3
b. x²+4x-8=0

Answer 1

A. x^2 - 2x = 3
=> x^2 - 2x - 3 = 0
a = 1, b = -2, c = -3
x = (-b ± √(b^2 - 4ac))/2a
x = (2 ± √(4 + 12)) / 2
x = (2 ± √16) / 2
x = (2 ± 4)/2
x = 1 ± 2
x = 1 + 2 = 3 atau x = 1 - 2 = -1

b. x^2 + 4x - 8 = 0
a = 1, b = 4 , c = -8
x = (-b ± √(b^2 - 4ac)) / 2a
x = (-4 ± √(16 + 32)) / 2
x = (-4 ± √48) / 2
x = (-4 ± 4√3) / 2
x = (-2 ± 2√3)
x = -2 + 2√3 atau x = -2 - 2√3