If the roots of the equation $12x^2+mx+5-0$ are in the ratio 3 : 2, then m equals

Let the roots be 3k and 2k, where k is a common multiplier. By Vieta's formulas, the product of the roots is 3k * 2k = 6k^2 = 5/12, which yields k^2 = 5/72 and k = sqrt(10)/12. The sum of the roots is 3k + 2k = 5k = -m/12. Substituting k gives -m/12 = 5*sqrt(10)/12, so m = -5*sqrt(10)/3, which is not among the listed numerical values.

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