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NathanSoares1
@NathanSoares1
December 2019
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Resolva a seguinte equação do 2° grau: -3x² = 2x-1
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fernandorioluz
-3x² = 2x - 1
-3x² -2x + 1 = 0
aplicando Bhaskara temos:
x = -b +/- √Δ / 2a, onde Δ=b²-4ac⇒Δ=(-2)²-4(-3).1⇒Δ=4+12= 16
x = -(-2) +/- √16 / 2.(-3)
x = 2 +/- 4 / -6
x1 = 2 -4 / -6 = -2/-6 = 1/3
x2 = 2 + 4 /-6 = 6/-6 = -1
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-3x² -2x + 1 = 0
aplicando Bhaskara temos:
x = -b +/- √Δ / 2a, onde Δ=b²-4ac⇒Δ=(-2)²-4(-3).1⇒Δ=4+12= 16
x = -(-2) +/- √16 / 2.(-3)
x = 2 +/- 4 / -6
x1 = 2 -4 / -6 = -2/-6 = 1/3
x2 = 2 + 4 /-6 = 6/-6 = -1