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Delazeribeta
@Delazeribeta
December 2019
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Determine os valores reais de x para que se tenha (x²+2)²=2(x²+6)
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mfmateus31
(x²+2)²=2(x²+6)
x^4+4x²+4=2x²+12
x^4+4x²+4 -2x²-12 = 0
x^4+2x²-8 = 0
x²=y
y²+2y-8=0
Δ=2²-4*1*-8
Δ= 4+32
Δ= 36
y= (-2+-√32)/2*1
y= (-2+-5,35)/2
y'=(-2+5,35)/2
y'=3,35/2
y'=1,675
y"=(-2-5,35)/2
y"=-7,35/2
y"= -3,675
x²=y
x=√1,675=1,294
x=√-3,675
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x^4+4x²+4=2x²+12
x^4+4x²+4 -2x²-12 = 0
x^4+2x²-8 = 0
x²=y
y²+2y-8=0
Δ=2²-4*1*-8
Δ= 4+32
Δ= 36
y= (-2+-√32)/2*1
y= (-2+-5,35)/2
y'=(-2+5,35)/2
y'=3,35/2
y'=1,675
y"=(-2-5,35)/2
y"=-7,35/2
y"= -3,675
x²=y
x=√1,675=1,294
x=√-3,675