Quadratic equation with one of the roots as a part of equation

Question

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

Hint: if the roots of a quadratic $p(x)$ are $a,b$ what are the roots of $x^2p\left(\frac 1x\right)$? — lulu, Jun 05 '21 at 16:02
@lulu It would be $\frac 1 a$ and $\frac 1 b$, but how do I apply that here? — Righter, Jun 05 '21 at 16:04
Well, first write down the quadratic with roots $4-\alpha, 4-\beta$ and then use my hint. — lulu, Jun 05 '21 at 16:05
Simplifies things a bit to note that, if $p(x)=Ax^2+Bx+C$ then $x^2p\left(\frac 1x\right)=Cx^2+Bx+A$ — lulu, Jun 05 '21 at 16:06
@lulu Well, yes, but after that aren't we still left with $\alpha$ terms? — Righter, Jun 05 '21 at 16:17
@LalitTolani The sum of the roots is zero. The term is $-4\alpha$, not $-4\alpha x$ — Righter, Jun 05 '21 at 16:17
Of course there are terms that depend on $\alpha$. $\alpha $ is not determined by the data. It almost is...of course $\alpha =2 \pm \sqrt 3$. But we don't know which of those two $\alpha $ is, so the final answer will have $\alpha's $ in it. — lulu, Jun 05 '21 at 16:18

score 1 · Accepted Answer · answered Jun 05 '21 at 16:52

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.

Quadratic equation with one of the roots as a part of equation

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