<s>[INST] Finding the equation whose roots are Okay I found this one on a test and it must be something really silly I am missing $ \alpha$ is not equal to $\beta$ and $\alpha^2=5\alpha-3,5\beta-3=\beta^2$ then the equation whose roots are $\alpha/\beta$ and $ \beta/\alpha$ is $ (a)3x^2-25+3=0,(b)x^2-5x+3=0,(c)x^2+5x-3=0,(d)3x^2-19x+3=0$ Of course the answer is either a or d I tried solving for $\alpha$ and $\beta$ but the equation became too complicated 
 [/INST] Notice, the roots $\alpha$ & $\beta$ are distinct roots &
$$\alpha^2=5\alpha -3, \ \  5\beta-3=\beta^2$$
Hence, the quadratic equation is $$x^2-5x+3=0$$
whose roots are determined by using quadratic formula $$\alpha, \beta=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-3)}}{2(1)}$$ $$=\frac{5\pm\sqrt{37}}{2}$$ $$\implies \alpha=\frac{5+\sqrt{37}}{2}, \ \ \beta=\frac{5-\sqrt{37}}{2}$$
Now, $$\frac{\alpha}{\beta}=\frac{\frac{5+\sqrt{37}}{2}}{\frac{5-\sqrt{37}}{2}}=-\frac{31+5\sqrt {37}}{6}$$
$$\frac{\beta}{\alpha}=\frac{\frac{5-\sqrt{37}}{2}}{\frac{5+\sqrt{37}}{2}}=-\frac{31+5\sqrt {37}}{6}=-\frac{31-5\sqrt {37}}{6}$$
Hence, the quadratic equation having roots $\frac{\alpha}{\beta}$ & $\frac{\beta}{\alpha}$ is given as $$\left(x-\frac{\alpha}{\beta}\right)\left(x-\frac{\beta}{\alpha}\right)=0$$
$$\left(x+\frac{31+5\sqrt {37}}{6}\right)\left(x+\frac{31-5\sqrt {37}}{6}\right)=0$$
$$\color{red}{3x^2+31x+3=0}$$
</s>