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(A) Let the roots of the quadratic equation $12x^2 + mx + 5 = 0$ be $3\alpha$ and $2\alpha$.According to the relationship between roots and coefficien

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$5\sqrt{10}$

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(A) Let the roots of the quadratic equation $12x^2 + mx + 5 = 0$ be $3\alpha$ and $2\alpha$.According to the relationship between roots and coefficients:Sum of roots: $3\alpha + 2\alpha = -\frac{m}{12} \Rightarrow 5\alpha = -\frac{m}{12} \Rightarrow \alpha = -\frac{m}{60}$.Product of roots: $(3\alpha)(2\alpha) = \frac{5}{12} \Rightarrow 6\alpha^2 = \frac{5}{12} \Rightarrow \alpha^2 = \frac{5}{72}$.Substituting the value of $\alpha$ into the product equation:$(-\frac{m}{60})^2 = \frac{5}{72} \Rightarrow \frac{m^2}{3600} = \frac{5}{72}$.$m^2 = \frac{3600 	imes 5}{72} = 50 	imes 5 = 250$.$m = \sqrt{250} = 5\sqrt{10}$ (since $m$ must be positive).

The positive value of $m$ for which the roots of the equation $12x^2 + mx + 5 = 0$ are in the ratio $3:2$ is

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