The roots of $2x^2 − 8x − 1$ are $\alpha$ and $\beta$. Suppose another quadratic, $x^2 + qx + r$, has roots $1/(\alpha^3\beta)$ and $1/(\beta^3\alpha)$. What are $q$ and $r$?
What I did to solve this question was $-q = (\beta^2 + \alpha^2)/(\alpha^3\beta^3)$
I sub in for the denominator $-1/2$, but I'm not sure what to do with the numerator?
Thank you.
On
The issue may be one of checking. I do get $q=136, r=16$ also and it seems to check out. Checking requires rationalizing the denominator, and a hidden trick lies there.
From the quadratic formula for the original quadratic equation we get
$\alpha=(4+3\sqrt{2})/2$
$\beta=(4-3\sqrt{2})/2$
For the quadratic equation derived in the problem, $x^2+136x+16=0$ the roots, purposely left unlabeled, are
$-68+48\sqrt{2}$
$-68-48\sqrt{2}$
We are to prove that $1/(\alpha^3\beta)$ and $1/(\alpha\beta^3)$ match the latter roots, allowing either order.
Look at $1/(\alpha^3\beta)$. A convenient way to proceed is to first render
$1/(\alpha^3\beta)=\beta^2/(\alpha^3\beta^3)=-8\beta^2$
where we have rendered $\alpha\beta=-1/2$ from the coefficients of the given equation, thereby we have rationalized the denominator. If we put in the value if $\beta$ given above we obtain
$\color{blue}{1/(\alpha^3\beta)=-68+48\sqrt{2}}$
That checks out as one of the target roots. The reader should now verify that
$\color{blue}{1/(\alpha\beta^3)=-68-48\sqrt{2}}$
by the same method, thus both roots are matched and the claimed values of $q$ and $r$ are correct.
Guide:
Notice that
$$\alpha^2 + \beta^2 = (\alpha + \beta)^2-2\alpha\beta$$
and you already know the sum and product of $\alpha$ and $\beta$.