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85chacha
@85chacha
May 2019
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Bonsoir, (1ère S) aidez-moi svp pour ce calcul, je suis désemparé:
Résoudre l'équation bicarrée:
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ramzi13
On pose x²=y alors : x^4 + 4x²-5 =0⇔ (x²)² +4x²-5 =0 (^4 puissance4)
⇔y²+4y-5=0
Δ=b²-4ac=4²-4(1)(-5)=16+20=36
y1=(-b+√Δ)/2a=(-4+√36)/2=(-4+6)/2=2/2=1
y2=(-b-√Δ)/2a=(-4-6)/2=-10/2=-5
x²=y ⇔x²=1 ⇔x=1 ou x=-1
x²=-5( x n'existe pas)
alors: les solutions de l'équation x^4+4x²-5=0 est:s={-1 ; 1}
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⇔y²+4y-5=0
Δ=b²-4ac=4²-4(1)(-5)=16+20=36
y1=(-b+√Δ)/2a=(-4+√36)/2=(-4+6)/2=2/2=1
y2=(-b-√Δ)/2a=(-4-6)/2=-10/2=-5
x²=y ⇔x²=1 ⇔x=1 ou x=-1
x²=-5( x n'existe pas)
alors: les solutions de l'équation x^4+4x²-5=0 est:s={-1 ; 1}