Roots in a given ratio: Find the positive value of m such that the roots of 12x^2 + m x + 5 = 0 are in the ratio 3 : 2.

Introduction / Context:When roots are in a known ratio, we can parametrize them as 3t and 2t. Using Vieta’s formulas for a quadratic ax^2 + bx + c = 0 (sum = −b/a, product = c/a), we can solve for t and then deduce the parameter m. The “positive value” clause guides the sign choice for t and hence m.

Given Data / Assumptions:

• Equation: 12x^2 + m x + 5 = 0
• Roots in ratio 3 : 2 ⇒ roots 3t and 2t
• Sum = −m/12; Product = 5/12

Concept / Approach:Set up: 3t + 2t = 5t = −m/12 and (3t)(2t) = 6t^2 = 5/12. First get t^2 from the product, then use the sum to find m. Finally select the sign of t that yields the required positive m (because m depends linearly on t with a negative coefficient in the sum relation).

Step-by-Step Solution:

Verification / Alternative check:With m = 5√10, sum = −m/12 < 0, matching 5t with t negative; product remains 5/12 > 0. The roots keep the 3:2 ratio in magnitude and are both negative, which is permissible; only m’s positivity was required.

Why Other Options Are Wrong:

• 5√10/2, 5/12, 12/5, 10√5: Do not satisfy both Vieta relations with the 3:2 ratio under the “m > 0” condition.

Common Pitfalls:Forgetting that the sign of m depends on the sign of t via the sum relation. Also, mixing “ratio 3:2” with “difference” of roots—use the ratio parametrization directly.

Final Answer:5√10