JEE Main 2002 — Mathematics Algebra

We have $\alpha^2=5 \alpha-3$α2=5α−3 $$\begin{aligned} & \Rightarrow \alpha^2-5 \alpha+3=0 \Rightarrow \alpha=\frac{5 \pm \sqrt{13}}{2} \text {. Similarly, } \beta^2=5 \beta-3 \Rightarrow \alpha=\frac{5 \pm \sqrt{13}}{2} \\ & \therefore \alpha=\frac{5+\sqrt{13}}{2} \text { and } \beta=\frac{5-\sqrt{13}}{2} \text { or vice - versa } \\ & \alpha^2+\beta^2=\frac{50+26}{4}=19 \& \alpha \beta=\frac{1}{4}(25-13)=3 \end{aligned}$$​⇒α2−5α+3=0⇒α=25±13​​. Similarly, β2=5β−3⇒α=25±13​​∴α=25+13​​ and β=25−13​​ or vice - versa α2+β2=450+26​=19&αβ=41​(25−13)=3​ Thus, the equation having $\frac{\alpha}{\beta} \& \frac{\beta}{\alpha}$βα​&αβ​ as its roots is $$x^2-x\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+\frac{\alpha \beta}{\alpha \beta}=0 \Rightarrow x^2-x\left(\frac{\alpha^2+\beta^2}{\alpha \beta}\right)+1=0 \text { or } 3 x^2-19 x+1=0$$x2−x(βα​+αβ​)+αβαβ​=0⇒x2−x(αβα2+β2​)+1=0 or 3x2−19x+1=0