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x1,2 = -b +- √(b² - 4ac)
2a
= 3 +- √{(-3)²-4(1)(1)}
2(1)
= 3 +- √(9-4)
2
= 3 +- √5
2
x1 = (3+√5)/2 dan x2 = (3-√5)/2
(x + x1)(x + x2) = 0
Jadi, x² - 3x + 1 = 0 ⇔ {x + (3+√5)/2}{x + (3-√5)/2}