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Raquelbangtanboys
@Raquelbangtanboys
December 2019
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Quero o calculo dessa equação 2 grau identifique os coeficientes e determine as raízes se existir
a) 4+x (x-4)=x
b) 4x ao quadrado - 4x+1
c) x ao quadrado - 4x-12=0
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francitorres20
A) 4+x(x-4)=x
4x-1+x^2-4x-x=0
x^2-x-16
Δ
=b^2-4ac
Δ=(-1)^2 . 4.1..(-16)
Δ=1+64
Δ=65
b)Δ=4x^2 - 4x+1
Δ=(-4)^2 - 4.4.1
Δ=16-16
Δ=0
c)x^2 - 4x-12
Δ=-4^2 . 4.1. (-12)
Δ=16+48
Δ=64
x=- (-4) +ou- √Δ /2.1
x1=4+8/2=6
x2=4-8/2=-2
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Raquelbangtanboys
valeu pela ajuda Obrigado
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4x-1+x^2-4x-x=0
x^2-x-16
Δ=b^2-4ac
Δ=(-1)^2 . 4.1..(-16)
Δ=1+64
Δ=65
b)Δ=4x^2 - 4x+1
Δ=(-4)^2 - 4.4.1
Δ=16-16
Δ=0
c)x^2 - 4x-12
Δ=-4^2 . 4.1. (-12)
Δ=16+48
Δ=64
x=- (-4) +ou- √Δ /2.1
x1=4+8/2=6
x2=4-8/2=-2