Suppose that $x + \frac{1}{x} = -1$.
What is the value of $x^{2016} + \frac{1}{x^{2016}}$?
(A) 1
(B) 2
(C) $\frac{2\pi}{3}$
(D) $\frac{4\pi}{3}$
Worked Solution & Example Answer:Suppose that $x + \frac{1}{x} = -1$ - HSC - SSCE Mathematics Extension 2 - Question 10 - 2016 - Paper 1
Step 1
Solve for x
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Answer
Given the equation:
[x + \frac{1}{x} = -1]
Multiply through by x:
[x^2 + 1 = -x]
Rearranging gives:
[x^2 + x + 1 = 0]
Using the quadratic formula:
[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2}]
This simplifies to:
[x = \frac{-1 \pm i\sqrt{3}}{2}]
Step 2
Calculate x^n and 1/x^n
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Answer
From the results of x, we can rewrite to find x2016 and x20161:
Using properties of even powers and complex numbers, we calculate:
[\text{Let } z = \frac{1}{2} + \frac{i\sqrt{3}}{2} \text{, then } z^{2016} + \frac{1}{z^{2016}} = 2 \cdot \text{Re}(z^{2016}) = 2.]
