If \(x + \dfrac{1}{x} = 17\), then the value of \(x^2 + \dfrac{1}{x^2}\) is:
287
Use the identity \(\left(x + \dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} + 2\).
Rearranging gives \(x^2 + \dfrac{1}{x^2} = \left(x + \dfrac{1}{x}\right)^2 - 2\).
Substitute \(x + \dfrac{1}{x} = 17\): \(x^2 + \dfrac{1}{x^2} = 17^2 - 2\).
So \(x^2 + \dfrac{1}{x^2} = 289 - 2 = 287\).
Hence, the value of \(x^2 + \dfrac{1}{x^2}\) is 287.
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