If $\alpha,\beta$ are roots of $x^2-4\alpha+1=0,$ then equation whose roots are $\frac 1 {4-\alpha}, \frac 1 {4-\beta}$ is?
I'm not sure how to go about this. One thing I noticed is that $$\alpha^2 -4\alpha+1=0$$ $$1=\alpha(4-\alpha)$$ $$\alpha=\frac 1 {4-\alpha}$$ which we can use in the equation we need. But I'm not sure how to get a similar expression for $\beta$. Can anyone please help? Thanks
See given equation is $x^2-4\alpha+1=0$
Now we write something as $( )^2-4\alpha+1=0$.
Inside the bracket we write such an expression of $x$ such that substituting $x=\frac{1}{4-\alpha}$ in that expression leaves us only with $\alpha$ and our uppermost equation gets satisfied. Such an expression is $\displaystyle\frac{1}{\frac{-1}{x}+4}$. Therefore our required equation is
$\displaystyle{{(\frac{1}{\frac{-1}{x}+4}})^2-4\alpha+1}=0$
You may now simplify it
Note: We didn't discussed about another root $\frac{1}{4-\beta}$ as it s similar looking to $\frac{1}{4-\alpha}$. therefore if our equation works for one it works for other as well.