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(B) Given the quadratic equation $x^2 - 6x + a = 0$.Since $\alpha$ and $\beta$ are the roots,the sum of roots is $\alpha + \beta = -(-6)/1 = 6$.We are

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$8$

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(B) Given the quadratic equation $x^2 - 6x + a = 0$.Since $\alpha$ and $\beta$ are the roots,the sum of roots is $\alpha + \beta = -(-6)/1 = 6$.We are given the relation $3\alpha + 2\beta = 16$.We can rewrite this as $2(\alpha + \beta) + \alpha = 16$.Substituting the value of $\alpha + \beta = 6$ into the equation:$2(6) + \alpha = 16$$12 + \alpha = 16$$\alpha = 4$.Since $\alpha$ is a root of the equation,it must satisfy $x^2 - 6x + a = 0$:$(4)^2 - 6(4) + a = 0$$16 - 24 + a = 0$$-8 + a = 0$$a = 8$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 6x + a = 0$ and satisfy the relation $3\alpha + 2\beta = 16$,then the value of $a$ is

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