wyznacz pierwiastki wielomianu
a)f(x)=x2-2x-48
b)g(x)=(x+3)(2x-1)+(3x-1)(2x-1)
a
delta=4+192=196
x1=(2-14)/2=-6
x2=(2+14)/2=8
b
(x+3)(2x-1)+(3x-1)(2x-1)=(2x-1)(4x+2)
2x-1=0 4x+2=0
x=0,5 x=-0,5
a) x²-2x-48
Δ=4+192=196
√Δ=14
x₁=2-14 / 2 = -12/2= -6
x₂=2+14 / 2 = 16/2= 8
b) (x+3)(2x-1)+(3x-1)(2x-1)
2x²-x+6x-3+6x²-3x-2x+1=0
8x²-2=0
8x²=2
x²=1/4
x= -1/2 v x= 1/2
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a
delta=4+192=196
x1=(2-14)/2=-6
x2=(2+14)/2=8
b
(x+3)(2x-1)+(3x-1)(2x-1)=(2x-1)(4x+2)
2x-1=0 4x+2=0
x=0,5 x=-0,5
a) x²-2x-48
Δ=4+192=196
√Δ=14
x₁=2-14 / 2 = -12/2= -6
x₂=2+14 / 2 = 16/2= 8
b) (x+3)(2x-1)+(3x-1)(2x-1)
2x²-x+6x-3+6x²-3x-2x+1=0
8x²-2=0
8x²=2
x²=1/4
x= -1/2 v x= 1/2