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(C) Given the biquadratic equation $x^4-9x^3+31x^2-49x+30=0$. Since the coefficients are real,complex roots occur in conjugate pairs. Thus,if $(2-i)$

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(C) Given the biquadratic equation $x^4-9x^3+31x^2-49x+30=0$. Since the coefficients are real,complex roots occur in conjugate pairs. Thus,if $(2-i)$ is a root,then $(2+i)$ is also a root. Let the four roots be $\alpha, \beta, (2-i),$ and $(2+i)$. The sum of the roots is $\alpha + \beta + (2-i) + (2+i) = -(-9)/1 = 9$. $\alpha + \beta + 4 = 9 \implies \alpha + \beta = 5$. The product of the roots is $\alpha \cdot \beta \cdot (2-i)(2+i) = 30/1 = 30$. Since $(2-i)(2+i) = 2^2 - i^2 = 4 + 1 = 5$,we have $\alpha \cdot \beta \cdot 5 = 30 \implies \alpha \cdot \beta = 6$. We have $\alpha + \beta = 5$ and $\alpha \cdot \beta = 6$. Solving these,we get $\alpha = 2$ and $\beta = 3$ (since $\alpha Therefore,$2\alpha - \beta = 2(2) - 3 = 4 - 3 = 1$.

If $(2-i)$ is one of the roots of the equation $x^4-9x^3+31x^2-49x+30=0$ and $\alpha, \beta$ $(\alpha < \beta)$ are its real roots,then $2\alpha-\beta=$

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