If α ≠ β but α^2=5 α-3 and β^2=5 β-3 then the equation having α / β...

We have $\alpha^2=5 \alpha-3$$\begin{aligned}& \Rightarrow \alpha^2-5 \alpha+3=0 \Rightarrow \alpha=\frac{5 \pm \sqrt{13}}{2} \text{. Similarly, } \beta^2=5 \beta-3 \Rightarrow \alpha=\frac{5 \pm \sqrt{13}}{2}\\ & \therefore \alpha=\frac{5+\sqrt{13}}{2} \text{\text{ and } } \beta=\frac{5-\sqrt{13}}{2} \text{ or vice - versa }\\ & \alpha^2+\beta^2=\frac{50+26}{4}=19 & \alpha \beta=\frac{1}{4}(25-13)=3 \end{aligned}$$Thus,$ the equation having\frac{\alpha}{\beta} \& \frac{\beta}{\alpha}$as its roots is x²-x$\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+\frac{\alpha \beta}{\alpha \beta}=$0 \Rightarrow x²-x\left(\frac{\alpha^2+\beta^2}{\alpha \beta}\right)+$1=0 \text{ or }3 x²-19 x+1=0