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jika D > 0 , maka memiliki akar-akar real berbeda
jika D = 0, maka memiliki akar real kembar
jika D < 0 , maka memiliki akar-akar imajiner
2014x² - 2015 = 0
a = 2014
b = 0
c = -2015
b² - 4ac = 0² - 4(2014)(-2015)
= 4(2014)(2015)
4(2014)(2015) > 0, maka akar-akarnya real berbeda
penyelesaian :
2014x² - 2015 = 0
2014x² = 2015
x = 2015/2014
x = √(2015/2014) atau -√(2015/2014)
√2x² - x - 1 = 0
a = √2
b = -1
c = -1
b² - 4ac = (-1)² - 4(√2)(-1)
= 1 + 4√2
1 + 4√2 > 0, maka akar-akarnya real berbeda
penyelesaian :
x1 = (-b + √D)/(2a)
= (-(-1) + (1 + 4√2))/(2√2)
= (2 + 4√2)/(2√2)
= 1/√2 + 2
x2 = (-b - √D)/(2a)
= (-(-1) - (1 + 4√2))/(2√2)
= (-4√2)/(2√2)
= -2
x² - 2016x = 0
a = 1
b = -2016
c = 0
b² - 4ac = (-2016)² - 4(1)(0)
= 2016²
2016² > 0, maka akar-akarnya real berbeda
penyelesaian :
x² - 2016x = 0
x(x - 2016) = 0
x = 0 atau x = 2016