Wielomian W(x)=9( zapisz w postaci iloczynu czterech wielomianow pierwszego stopnia
w(x)=(9x²-8)(2x²-x-10)
(9x²-8)=0
Δ=0-4*9*(-8)=-36*(-8)= 288
√Δ=2√(72)= 2√(36*2)=12√2
x1=-12√2/18=(-2√2)/3
x2=(2√2)/3
(2x²-x-10)=0
Δ=1-4*2*(-10)=1-8*(-10)= 81
√Δ=9
x1=(1-9)/4=-8/4=-2
x2=(1+9)/4=10/4=5/2
w(x)=(9x²-8)(2x²-x-10) =
=(3x-2√2)(3x+2√2)(x+2)(2x-5)
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w(x)=(9x²-8)(2x²-x-10)
(9x²-8)=0
Δ=0-4*9*(-8)=-36*(-8)= 288
√Δ=2√(72)= 2√(36*2)=12√2
x1=-12√2/18=(-2√2)/3
x2=(2√2)/3
(2x²-x-10)=0
Δ=1-4*2*(-10)=1-8*(-10)= 81
√Δ=9
x1=(1-9)/4=-8/4=-2
x2=(1+9)/4=10/4=5/2
w(x)=(9x²-8)(2x²-x-10) =
=(3x-2√2)(3x+2√2)(x+2)(2x-5)
licze na naj