If $ x + \frac{1}{x} = \frac{k}{2}$, then what is the value of $\frac{x^8+1}{x^4}$ ?
Answer & explanation
Correct answer: option 1
If $ x + \frac{1}{x} = \frac{k}{2}$,
then what is the value of $\frac{x^8+1}{x^4}$
We can write $\frac{x^8+1}{x^4}$ as x4 + \(\frac{1}{x^4}\)
If $(x + \frac{1}{x}) = x
(x2 + x-2) = (x)2 - 2 = b
= (x4 + x-4) = b2 - 2
$ x + \frac{1}{x} = \frac{k}{2}$,
(x2 + x-2) = ($ \frac{k}{2}$)2 - 2 = \(\frac{k^2 - 8}{4}\)
and, x4 + \(\frac{1}{x^4}\)= (\(\frac{k^2 - 8}{4}\))2 - 2
x4 + \(\frac{1}{x^4}\) = $\frac{k^4-16k^2+32}{16}$