Zadanie 6
a)
x^3-6x^2+8x=0
x(x^2-6x+8)=0 , -6x=-2x-4x
x(x^2-2x-4x+8)=0
x[x(x-2)-4(x-2)]=0
x(x-2)(x-4)=0
x=0 \ \vee \ x=2 \ \vee \ x=4
x\in \{0, \ 2, \ 4\}
b)
x^3-4x^2-12x=0
x(x^2-4x-12)=0 , -4x=2x-6x
x(x^2+2x-6x-12)=0
x[x(x+2)-6(x+2)]=0
x(x+2)(x-6)=0
x=0 \ \vee \ x=-2 \ \vee \ x=6
x\in \{-2,\ 0, \ 6\}
c)
4x^3-7x^2-2x=0
x(4x^2-7x-2)=0 , -7x=-8x+x
x(4x^2-8x+x-2)=0
x[4x(x-2)+(x-2)]=0
x(x-2)(4x+1)=0
x=0 \ \vee \ x = 2 \ \vee \ 4x=-1 \ \Rightarrow \ x=-\frac{1}{4}
x\in \{-\frac{1}{4}, \ 0, \ 2\}
d)
2x^3 +4x^2 +x=0
x(2x^2+4x+1)=0
x=0
lub
2x^2+4x+1=0
a=2, b=4, c=1
\Delta=4^2-4\cdot 2\cdot 1=8
\sqrt\Delta=\sqrt{4\cdot 2}=2\sqrt2
x_1=\frac{-4-2\sqrt2}{2\cdot 2}=\frac{-4-2\sqrt2}{4}=-1-\frac{\sqrt2}{2}
x_2=\frac{-4+2\sqrt2}{2\cdot 2}=\frac{-4+2\sqrt2}{4}=\frac{\sqrt2}{2}-1
x\in \{-1-\frac{\sqrt2}{2} \ , \ \frac{\sqrt2}{2}-1\}
Zastosowany wzór skróconego mnożenia
(a+b)^2=a^2+2ab+b^2