MHB Value of $\dfrac{2k^2}{k-1}$: Solving the Equation

[anemone]

Determine the value of $\dfrac{2k^2}{k-1}$ given $\dfrac{k^2}{k-1}=k^2-8$.

 

[anemone]

Solution of other:

Spoiler

Rewrite the given equation as a cubic equation, we see that:

$k^3-2k^2-8k+8=0$

Let $k=\dfrac{2+4\sqrt{7}x}{3}$, the above cubic equation becomes $4x^4-3x=-\dfrac{1}{2\sqrt{7}}$.

Let $x=\cos \theta$, $4x^4-3x=-\dfrac{1}{2\sqrt{7}}$ translates into $\cos 3\theta=-\dfrac{1}{2\sqrt{7}}$ and this gives

$x=\cos\left(\dfrac{1}{3}\arccos \left(-\dfrac{1}{2\sqrt{7}}\right)+\dfrac{2n\pi}{3}\right)$, where $n\in\{0,\,1,\,2\}$

So

$\dfrac{2k^2}{k-1}=\dfrac{8(28\cos^2\left(\dfrac{1}{3}\arccos \left(-\dfrac{1}{2\sqrt{7}}\right)+\dfrac{2n\pi}{3}\right)+4\sqrt{7}\cos \left(\dfrac{1}{3}\arccos \left(-\dfrac{1}{2\sqrt{7}}\right)+\dfrac{2n\pi}{3}\right)+1}{3\left(4\sqrt{7}\cos \left(\dfrac{1}{3}\arccos \left(-\dfrac{1}{2\sqrt{7}}\right)+\dfrac{2n\pi}{3}\right)-1 \right)}$,

where $n\in\{0,\,1,\,2\}$

Compute the value of the targeted expression with calculator, by first taking $n=0$ and then replace it by $n=1$ and next $n=2$ we get three different values for $\dfrac{2k^2}{k-1}$:

$9.9758,\,-3.5603,\,-14.4155$

 

[Albert1]

Spoiler

we have :$k^3-2k^2-8k+8=0-----(1)$
to find the solution of (1) we can use Newton's method
the roots of (1)
$k=-2.494, 0.8901, 3.609$
$\therefore \dfrac {2k^2}{k-1}=-3.56041,-14.4182, 9.975879$