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(B) Let $\alpha$ be the common root of the equations $x^2 - 5ax + 1 = 0$ and $x^2 - ax - 5 = 0$.Then $\alpha^2 - 5a\alpha + 1 = 0$ and $\alpha^2 - a\a

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$13$

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(B) Let $\alpha$ be the common root of the equations $x^2 - 5ax + 1 = 0$ and $x^2 - ax - 5 = 0$.Then $\alpha^2 - 5a\alpha + 1 = 0$ and $\alpha^2 - a\alpha - 5 = 0$.Subtracting the two equations: $(\alpha^2 - 5a\alpha + 1) - (\alpha^2 - a\alpha - 5) = 0$.$-4a\alpha + 6 = 0 \Rightarrow \alpha = \frac{6}{4a} = \frac{3}{2a}$.Substitute $\alpha = \frac{3}{2a}$ into $x^2 - ax - 5 = 0$:$(\frac{3}{2a})^2 - a(\frac{3}{2a}) - 5 = 0$.$\frac{9}{4a^2} - \frac{3}{2} - 5 = 0$.$\frac{9}{4a^2} = \frac{13}{2}$.$a^2 = \frac{9 	imes 2}{4 	imes 13} = \frac{9}{26}$.Since $a > 0$,$a = \frac{3}{\sqrt{26}}$.Given $a = \frac{3}{\sqrt{2\beta}}$,we have $\sqrt{2\beta} = \sqrt{26}$,so $2\beta = 26$,which means $\beta = 13$.

If the value of real number $a > 0$ for which $x^2 - 5ax + 1 = 0$ and $x^2 - ax - 5 = 0$ have a common real root is $\frac{3}{\sqrt{2\beta}}$,then $\beta$ is equal to

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