Question

biquadradas x²(x²-1)=2(x²-1)+6
pfvv​

Answer 1

x⁴-x²=2x²-2+6

x⁴-x²-2x²-6=0

x⁴-3x²-6= 0

x²=y

x⁴=y²

y²-3y-6=0

∆ = (-3)²-4(1)(-4)

∆ = 9 + 16 = 25

x= 3+-5/2

x¹= 3+5/2 = 4

x² = 3-5/2 = -1

x²= y

x²= 4

x = √4 = -2/2

x = √-1// n existe

S = {-2,2}