Discussion
If \( x + \dfrac{1}{x} = 5 \), what is the value of \( x^{2} + \dfrac{1}{x^{2}} \)?
Answer: A
Correct answer: 23
Square both sides of the given equation: \[ \left(x + \frac{1}{x}\right)^{2} = x^{2} + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^{2}} = x^{2} + \frac{1}{x^{2}} + 2. \] So \[ 25 = x^{2} + \frac{1}{x^{2}} + 2 \;\Rightarrow\; x^{2} + \frac{1}{x^{2}} = 23. \] The middle term is always exactly 2 because \( x \) and \( 1/x \) multiply to 1, and that is what makes the family of identities so quick to use. The related results are worth having ready. \( \left(x - \frac{1}{x}\right)^{2} = x^{2} + \frac{1}{x^{2}} - 2 \), so here \( x - \frac{1}{x} = \sqrt{21} \). And \( x^{3} + \frac{1}{x^{3}} = \left(x + \frac{1}{x}\right)^{3} - 3\left(x + \frac{1}{x}\right) = 125 - 15 = 110 \). The distractor 25 is the square of the given value, offered for a reader who forgets to subtract the 2.
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