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

Given,

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

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

${k}^{2} - 5\alphak + 1 = 0 . . . . . . \left(1\right)</p><p>${k}^{2} - \alphak - 5 = 0 . . . . . \left(2\right)$</p><p>Now subtraction above equations we get,</p><p>$k = \dfrac{6}{4\alpha}$</p><p>Now putting the value of$k$in equation \left(1\right)$ we get,

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

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

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

$⇒ {\alpha}^{2} = \dfrac{9}{26}$

$⇒ \alpha = \dfrac{3}{\sqrt{2 \times 13}}$

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