If α , β are the roots of the equation x 2 - 5 + 3 log 3 5 - 5 log 5 3 x + 3 3 log 3 5 1 3 - 5 log…

Given $\alpha ,\beta$ are the roots of the equation $x^2-\left(5+3^{\sqrt{{\log }_{3}5}}-5^{\sqrt{{\log }_{5}3}}\right)x+3\left(3^{{\left({\log }_{3}5\right)}^{\frac{1}{3}}}-5^{{\left({\log }_{5}3\right)}^{\frac{2}{3}}}-1\right)=0$

Now $3^{\sqrt{{\log }_{3}5}}-5^{\sqrt{{\log }_{5}3}}=3^{\sqrt{{\log }_{3}5}}-{\left(3^{{\log }_{3}5}\right)}^{\sqrt{{\log }_{5}3}}$

$=3^{\sqrt{{\log }_{3}5}}-3^{\sqrt{{\log }_{3}5}}=0$

Also $3^{{\left({\log }_{3}5\right)}^{\frac{1}{3}}}-5^{{\left({\log }_{5}3\right)}^{\frac{2}{3}}}=5^{{\left({\log }_{5}3\right)}^{\frac{2}{3}}}-5^{{\left({\log }_{5}3\right)}^{\frac{2}{3}}}$

$=0$

So, given equation becomes $x^2-5x-3=0$ 

i.e. $\alpha +\beta =5; \alpha \beta =-3$

Now if the roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$ i.e., $\frac{\alpha \beta + 1}{\beta}$ and $\frac{\alpha \beta + 1}{\alpha}$ i.e., $-\frac{2}{\beta}$ and $-\frac{2}{\alpha}$ Then let $-\frac{2}{\alpha} = t so \alpha = -\frac{2}{t}$ As $\alpha^2 - 5\alpha - 3 = 0$ Then $(-\frac{2}{t})^2 - 5(-\frac{2}{t}) - 3 = 0$ Which simplifies to $\frac{4}{t^2} + \frac{10}{t} - 3 = 0$ And further simplifies to $3t^2 - 10t - 4 = 0$ i.e., $3x^2 - 10x - 4 = 0$ is the required equation.