Mathematics - Quadratic Equation Question with Solution | TestHub

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., $\frac{\alpha \beta +1}{\beta }\&\frac{\alpha \beta +1}{\alpha }$ 

i.e., $\frac{-2}{\beta } \& \frac{-2}{\alpha }$

Then let $\frac{-2}{\alpha }=t\Rightarrow \alpha =\frac{-2}{t}$

As ${\alpha }^2-5\alpha -3=0$

$\Rightarrow {\left(\frac{-2}{t}\right)}^2-5\left(\frac{-2}{t}\right)-3=0$

$\Rightarrow \frac{4}{t^2}+\frac{10}{t}-3=0$

$\Rightarrow 3t^2-10t-4=0$

i.e., $3x^2-10x-4=0$ is the required equation.