Mathematics - Quadratic Equation Question with Solution | TestHub

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