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(A) Given equation: $x^5-5 x^4+9 x^3-9 x^2+5 x-1=0$. $x=1$ is a root of the equation. Dividing by $(x-1)$,we get $(x-1)(x^4-4 x^3+5 x^2-4 x+1)=0$. For

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$-1, \frac{1}{3}$

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(A) Given equation: $x^5-5 x^4+9 x^3-9 x^2+5 x-1=0$. $x=1$ is a root of the equation. Dividing by $(x-1)$,we get $(x-1)(x^4-4 x^3+5 x^2-4 x+1)=0$. For $x^4-4 x^3+5 x^2-4 x+1=0$,divide by $x^2$: $x^2-4 x+5-\frac{4}{x}+\frac{1}{x^2}=0 $ $\Rightarrow (x^2+\frac{1}{x^2})-4(x+\frac{1}{x})+5=0 $ $\Rightarrow (x+\frac{1}{x})^2-2-4(x+\frac{1}{x})+5=0 $ $\Rightarrow (x+\frac{1}{x})^2-4(x+\frac{1}{x})+3=0$. Let $y = x+\frac{1}{x}$,then $y^2-4y+3=0 $ $\Rightarrow (y-1)(y-3)=0 $ $\Rightarrow y=1, 3$. Case $1$: $x+\frac{1}{x}=1 \Rightarrow x^2-x+1=0$ (roots are complex). Case $2$: $x+\frac{1}{x}=3 \Rightarrow x^2-3x+1=0$ (roots are irrational). Thus,$\alpha+\beta=3$ and $\alpha \beta=1$. Substituting into $(\alpha+\beta) x^2+2 \alpha \beta x-\alpha \beta=0$: $3x^2+2x-1=0 $ $\Rightarrow (3x-1)(x+1)=0 $ $\Rightarrow x=-1, \frac{1}{3}$.

If $\alpha, \beta$ are the irrational roots of the equation $x^5-5 x^4+9 x^3-9 x^2+5 x-1=0$,then the roots of the equation $(\alpha+\beta) x^2+2 \alpha \beta x-\alpha \beta=0$ are

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