If the value of real number α > 0 for which x 2 - 5 α x + 1 = 0 and x 2 - α x - 5 = 0…

Given,

Quadratic equation $x^2-5\alpha x+1=0$  and $x^2-\alpha x-5=0$ have a common real root,

So, taking $k$ to be common root then equation becomes,

$k^2-5\alpha k+1=0 ......\left(1\right)$

$k^2-\alpha k-5=0 .....\left(2\right)$

Now subtraction above equations we get,

$k=\frac{6}{4\alpha }$

Now putting the value of $k$ in equation $\left(1\right)$ we get,

${\left(\frac{6}{4\alpha }\right)}^2-5\alpha \left(\frac{6}{4\alpha }\right)+1=0$

$\Rightarrow {\left(\frac{6}{4\alpha }\right)}^2-\left(\frac{26}{4}\right)=0$

$\Rightarrow \frac{36}{16{\alpha }^2}=\left(\frac{26}{4}\right)$

$\Rightarrow {\alpha }^2=\frac{9}{26}$

$\Rightarrow \alpha =\frac{3}{\sqrt{2\times 13}}$

Now comparing with given value $\frac{3}{\sqrt{2\beta }}$ we get, $\beta =13$