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.
 A: Let the new roots be $\gamma = \frac {\alpha}{\beta}$ and $\delta = \frac {\beta}{\alpha}$.
Compute $\gamma\delta =1$ and $\gamma+\delta = \frac {\alpha^2+\beta^2}{\alpha\beta}$
Note that $(\alpha+\beta)^2-2\alpha\beta =\alpha^2+\beta^2$
Can you put the pieces together now?
A: If $α$ and $β$ are the roots of the quadratic equation $ax^2+bx+c=0$ ,that is, $x^2+\frac{b}{a}x+\frac{c}{a}=0$ ($a$ is not equal to zero), then the equation can be written as $x^2-(α+β)x+αβ=0$.     
The equation whose roots are $\frac{α}{β}$ and $\frac{β}{α}$ is: $x^2-(\frac{α}{β}+\frac{β}{α})x+(\frac{α}{β})(\frac{β}{α})=0$.
$\implies x^2-(\frac{α^2+b^2}{αβ})x+1=0$
$\implies x^2-(\frac{(α+b)^2-2αb}{αβ})x+1=0$
$α$ and $β$ are the roots of the equation $x² + 8x - 5 = 0$
$α+β=-8$
$αβ=-5$
Therefore the required equation is $x^2-(\frac{(-8)^2-2(-5)}{-5})x+1=0$
$\implies x^2-(\frac{74}{-5})x+1=0$
$\implies x^2+\frac{74}{5}x+1=0$
$\implies 5x^2+74x+5=0$
A: $$(\alpha x-\beta)(\beta x-\alpha)=\alpha\beta x^2-(\alpha^2+\beta^2)x+\alpha\beta.$$
As $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$, using the Vieta formulas, the equation is
$$-5x^2-(8^2+2\cdot5)x-4=0,$$
$$5x^2+74x+5=0.$$
A: The product of the roots $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$ is $1$. That part was easy!  The sum will be more work.
The sum $\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha}$ of the roots simplifies to $\dfrac{\alpha^2+\beta^2}{\alpha\beta}$. 
But $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$.  Thus the sum of the roots is  $\dfrac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$.
Substituting the known values of $\alpha+\beta$ and $\alpha\beta$ we find that the sum of the roots is $\dfrac{(-8)^2-2(-5)}{-5}$. 
This simplifies to $-\dfrac{74}{5}$.
Thus the equation is 
$$x^2+\frac{74}{5}x+1=0.$$
  We can multiply through by $5$ if we wish.
A: 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: $$  \frac{\alpha}{\beta} \ + \ \frac{\beta}{\alpha} \ \ = \ \ \frac{\alpha^2 \ + \ \beta^2}{\alpha·\beta} \ \ = \ \ \frac{(-4 \ + \ \sqrt{21})^2 \ + \ (-4 \ - \ \sqrt{21})^2}{(-4 \ + \ \sqrt{21})·(-4 \ - \ \sqrt{21})} \ \ = \ \ \frac{2·[ \ (-4)^2 \ + \ (\sqrt{21})^2 \ ]}{(-4)^2 \ - \ (\sqrt{21})^2} $$ $$ = \ \ \frac{2·( 16 \ + \ 21 )}{16 \ - \ 21} \ = \ -\frac{74}{5} \ \ . $$
A: By transformation
Noting that $$
\frac{\alpha}{\beta}=\frac{\alpha \beta}{\beta^2}=-\frac{5}{\beta^2} \textrm{ and } \frac{\beta}{\alpha}=\frac{\alpha \beta}{\alpha^2}=-\frac{5}{\alpha^2},
$$ we transform the given equation by squaring it and then putting $y= -\frac{5}{x^2}$  as
$$
\begin{aligned} &(8 x)^2=(5-x^2)^2\\
\Rightarrow \quad  & 64 x^2=25-10 x^2+x^4 \\
\Rightarrow \quad  & 74\left(-\frac{5}{y}\right)=25+\left(-\frac{5}{y}\right)^2 \\
\Rightarrow \quad  & 5 y^2 +74 y +5=0, 
\end{aligned}
$$
which is the quadratic equation whose roots are $\frac{\alpha}{\beta} \textrm{ and } \frac{\beta}{\alpha}$.
