A) \[10\sqrt{5}\]
B) \[-5\sqrt{10}\]
C) 50
D) 25
Correct Answer: B
Solution :
Let roots of the equation\[12{{x}^{2}}+mx+5=0\] are \[\alpha \]and\[\beta ,\]when \[\alpha +\beta =-\frac{m}{12},\alpha \beta =\frac{5}{12}\] ?(i) But ratio in the roots is\[3:2\] Thus, \[\frac{\alpha }{\beta }=\frac{3}{2}\] \[\Rightarrow \] \[\alpha =\frac{3}{2}\beta \] From Eq. (i), \[\frac{3}{2}\beta +\beta =-\frac{m}{12}\] and \[\frac{3}{2}\beta .\beta =\frac{5}{12}\] \[\Rightarrow \] \[\frac{5}{2}\beta =-\frac{m}{12},{{\beta }^{2}}=\frac{5}{12}\times \frac{2}{3}\] \[\Rightarrow \] \[-m=12\times \frac{5}{2}\beta ,\beta =\sqrt{\frac{5}{18}}\] \[\therefore \] \[-m=30.\sqrt{\frac{5}{18}}=30.\sqrt{\frac{5\times 2}{9\times 2\times 2}}\] \[=30.\frac{\sqrt{10}}{3\times 2}\] \[-m=5\sqrt{10}\] \[\Rightarrow \] \[m=-5\sqrt{10}\]
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