If \(x + \dfrac{1}{x} = -1\), then compute: \(x^4 + \dfrac{1}{x^4} + 2x^2 + \dfrac{2}{x^2}\).
-3
Step 1 – find \(x^2 + \dfrac{1}{x^2}\) :
Square the given relation:
\(\left(x + \dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} + 2 = (-1)^2 = 1\)
So \(x^2 + \dfrac{1}{x^2} = 1 - 2 = -1\).
Step 2 – find \(x^4 + \dfrac{1}{x^4}\) :
\(\left(x^2 + \dfrac{1}{x^2}\right)^2 = x^4 + \dfrac{1}{x^4} + 2 = (-1)^2 = 1\)
So \(x^4 + \dfrac{1}{x^4} = 1 - 2 = -1\).
Step 3 – combine:
\(x^4 + \dfrac{1}{x^4} + 2\left(x^2 + \dfrac{1}{x^2}\right) = -1 + 2 \times (-1) = -1 - 2 = -3\).
Hence the answer is −3.
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