If the ratio of the roots of the equation 12x | vedclass

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(A) Let the roots be $\alpha$ and $\beta$. Given $\frac{\alpha}{\beta} = \frac{2}{3}$,so $\alpha = \frac{2\beta}{3}$.From the sum of roots,$\alpha + \

Vedclass · Accepted Answer

$5\sqrt{10}$

Vedclass · Answer

(A) Let the roots be $\alpha$ and $\beta$. Given $\frac{\alpha}{\beta} = \frac{2}{3}$,so $\alpha = \frac{2\beta}{3}$.From the sum of roots,$\alpha + \beta = \frac{m}{12}$ $\Rightarrow \frac{2\beta}{3} + \beta = \frac{m}{12}$ $\Rightarrow \frac{5\beta}{3} = \frac{m}{12}$ $\Rightarrow m = 5\beta 	imes 4 = 20\beta$.From the product of roots,$\alpha \beta = \frac{5}{12}$ $\Rightarrow \left(\frac{2\beta}{3}ight)\beta = \frac{5}{12}$ $\Rightarrow \frac{2\beta^2}{3} = \frac{5}{12}$ $\Rightarrow \beta^2 = \frac{5}{8}$ $\Rightarrow \beta = \sqrt{\frac{5}{8}} = \frac{\sqrt{5}}{2\sqrt{2}} = \frac{\sqrt{10}}{4}$.Substituting $\beta$ into the expression for $m$: $m = 20 	imes \frac{\sqrt{10}}{4} = 5\sqrt{10}$.

If the ratio of the roots of the equation $12x^2 - mx + 5 = 0$ is $2 : 3$,then $m = .....$

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