If $\alpha \neq \beta$ but $\alpha^2=5 \alpha-3$ and $\beta^2=5 \beta-3$ then the equation having $\alpha /…
If $\alpha \neq \beta$ but $\alpha^2=5 \alpha-3$ and $\beta^2=5 \beta-3$ then the equation having $\alpha / \beta$ and $\beta / \alpha$ as its roots is
$3 x^2-19 x+3=0$
$3 x^2+19 x-3=0$
$3 x^2-19 x-3=0$
$x^2-5 x+3=0$
Solution
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 { 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
$
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
$