<s>[INST] Quadratic equation - $\alpha$ and $\beta$ Roots If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 8x - 5 = 0$, find the quadratic equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.
My working out so far:
I know that $\alpha + \beta = -8$ and $\alpha \beta = -5$ (from the roots) and thenIi go on to work out that
$\alpha= -8-\beta$ and $\beta= -8-\alpha$, then I substitute into what the question asks me.
$\frac{-8-\beta}{-8-\alpha}$ and $\frac{-8-\alpha}{-8-\beta}$ however I do not know how to proceed further. I might be doing this completely wrong and my apologies for that.
Another solution came to me that if $\alpha$ and $\beta$ are roots of the other unknown equation. I can somehow manipulate that to find the answer. But I don't think that will work. All help is appreciated thank you.
 [/INST] Since we want the zeroes of the new quadratic polynomial to be the mutual reciprocals $ \ \frac{\alpha}{\beta} \ $ and $ \ \frac{\beta}{\alpha} \ \ , $ it must be palindromic, that is, of the form $ \ Ax^2 + Bx + A \ \  $ (the sort of symmetry André Nicolas was referring to in his comments).  Viete tells us that $ \ \alpha + \beta \ = \ -8 \ $ and that $ \ \alpha·\beta \ = \ -5 \ \ , $ as other posters have found.  So we may conclude that the sum of the reciprocals of these roots is
$$ \frac{\alpha \ + \ \beta}{\alpha · \beta} \ \ = \ \ \frac{1}{\alpha} \ + \ \frac{1}{\beta} \ \ = \ \ \frac{-8}{-5} \ \ = \ \ \frac85 \ \ . $$
We may also divide the sum-of-roots equation through to find that
$$   \frac{\alpha \ + \ \beta}{\alpha} \ \ = \ \ 1 \ + \ \frac{  \beta}{\alpha} \ \ = \ \ \frac{-8}{\alpha} \ \ \ \text{and} \ \ \ \frac{\alpha \ + \ \beta}{\beta} \ \ = \ \ 1 \ + \ \frac{\alpha}{ \beta} \ \ = \ \ \frac{-8}{\beta} \ \ . $$
We thus obtain
$$ \left(1 \ + \ \frac{  \beta}{\alpha} \right) \ + \ \left(1 \ + \ \frac{\alpha}{ \beta} \right) \  =  \ -8·\left(\frac{1}{\alpha} +  \frac{1}{\beta} \right) \ \ = \ \ -\frac{64}{5} $$ $$ \Rightarrow \ \ \frac{\alpha}{ \beta} \ + \ \frac{  \beta}{\alpha} \ \ = \ \ -\frac{64}{5} \ - \ 2 \ \ = \ \ -\frac{74}{5} \ \  . $$
Our monic palindromic polynomial is therefore $ \ x^2 \ + \  \frac{74}{5} · x \ + \ 1 \ \ . $  [In answer to the question, a corresponding quadratic equation is $ \ A·\left(x^2   +    \frac{74}{5} · x   +   1 \right) \ = \ 0 \ \ , $ where any non-zero value for $ \ A \ $ will do.]
$$ \ \ $$
It isn't overly difficult to work with the roots themselves:  the discriminant of the original quadratic polynomial is $ \ 8^2 + 4·5 \ = \ 84 \ \ , $ so the zeroes are $ \ \alpha \ , \ \beta \ = \ -4 \ \pm  \ \sqrt{21} \ \ . $  From this, we have reasonably convenient calculation with "conjugate" numbers: $$  \frac{\alpha}{\beta} \ + \ \frac{\beta}{\alpha} \ \ = \ \ \frac{\alpha^2 \ + \ \beta^2}{\alpha·\beta} \ \ = \ \ \frac{(-4 \ + \ \sqrt{21})^2 \ + \ (-4 \ - \ \sqrt{21})^2}{(-4 \ + \ \sqrt{21})·(-4 \ - \ \sqrt{21})} \ \ = \ \ \frac{2·[ \ (-4)^2 \ + \ (\sqrt{21})^2 \ ]}{(-4)^2 \ - \ (\sqrt{21})^2} $$ $$ = \ \ \frac{2·( 16 \ + \ 21 )}{16 \ - \ 21} \ = \ -\frac{74}{5} \ \ . $$
</s>