The number of all possible positive integral | vedclass

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(C) For the roots of the quadratic equation $ax^2 + bx + c = 0$ to be rational,the discriminant $D = b^2 - 4ac$ must be a perfect square.Here,$a = 6$,

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$3$

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(C) For the roots of the quadratic equation $ax^2 + bx + c = 0$ to be rational,the discriminant $D = b^2 - 4ac$ must be a perfect square.Here,$a = 6$,$b = -11$,and $c = \alpha$.$D = (-11)^2 - 4(6)(\alpha) = 121 - 24\alpha$.Since $\alpha$ is a positive integer,$121 - 24\alpha \ge 0$,which implies $24\alpha \le 121$,so $\alpha \le 5.04$. Thus,$\alpha \in \{1, 2, 3, 4, 5\}$.We check each value:If $\alpha = 1$,$D = 121 - 24(1) = 97$ (not a perfect square).If $\alpha = 2$,$D = 121 - 24(2) = 121 - 48 = 73$ (not a perfect square).If $\alpha = 3$,$D = 121 - 24(3) = 121 - 72 = 49 = 7^2$ (perfect square).If $\alpha = 4$,$D = 121 - 24(4) = 121 - 96 = 25 = 5^2$ (perfect square).If $\alpha = 5$,$D = 121 - 24(5) = 121 - 120 = 1 = 1^2$ (perfect square).The possible values for $\alpha$ are $3, 4, 5$. Therefore,there are $3$ such values.

The number of all possible positive integral values of $\alpha$ for which the roots of the quadratic equation $6x^2 - 11x + \alpha = 0$ are rational numbers is:

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