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(B) Given that $1+\sqrt{2}$ and $2-i$ are roots of the polynomial equation with rational coefficients,the conjugate roots $1-\sqrt{2}$ and $2+i$ must

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real and equal

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(B) Given that $1+\sqrt{2}$ and $2-i$ are roots of the polynomial equation with rational coefficients,the conjugate roots $1-\sqrt{2}$ and $2+i$ must also be roots.Let the roots be $\alpha_1 = 1+\sqrt{2}, \alpha_2 = 1-\sqrt{2}, \alpha_3 = 2-i, \alpha_4 = 2+i$.Using Vieta's formulas:$-b = \sum \alpha_i = (1+\sqrt{2}) + (1-\sqrt{2}) + (2-i) + (2+i) = 2 + 4 = 6 \Rightarrow b = -6$.$c = \sum \alpha_i \alpha_j = (1+\sqrt{2})(1-\sqrt{2}) + (1+\sqrt{2})(2-i) + (1+\sqrt{2})(2+i) + (1-\sqrt{2})(2-i) + (1-\sqrt{2})(2+i) + (2-i)(2+i) = -1 + 4 + 4 + 4 + 4 + 5 = 20$.Wait,recalculating $c$: $\alpha_1 \alpha_2 = -1$,$\alpha_3 \alpha_4 = 5$,$(\alpha_1+\alpha_2)(\alpha_3+\alpha_4) = 2 	imes 4 = 8$. So $c = -1 + 5 + 8 = 12$.$-d = \sum \alpha_i \alpha_j \alpha_k = \alpha_1 \alpha_2 (\alpha_3+\alpha_4) + \alpha_3 \alpha_4 (\alpha_1+\alpha_2) = -1(4) + 5(2) = -4 + 10 = 6$ $\Rightarrow d = -6$.The equation $bx^2+cx+d=0$ becomes $-6x^2+12x-6=0$,which simplifies to $x^2-2x+1=0$.This is $(x-1)^2=0$,so the roots are $1, 1$,which are real and equal.

If $1+\sqrt{2}$ and $2-i$ are the roots of the equation $x^4+bx^3+cx^2+dx+e=0$ where $b, c, d, e$ are rational numbers,then the roots of the equation $bx^2+cx+d=0$ are

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