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If $ x + \frac{1}{x} = \frac{k}{2}$, then what is the value of $\frac{x^8+1}{x^4}$ ? |
$\frac{k^4-16k^2+32}{16}$ $\frac{k^4-8k^2-36}{32}$ $\frac{k^4-8k^2-32}{16}$ $\frac{k^4-8k^2+32}{16}$ |
$\frac{k^4-16k^2+32}{16}$ |
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 + \(\frac{1}{x^4}\) = $\frac{k^4-16k^2+32}{16}$ |
