If $x - \dfrac{1}{x} = 4$, find $x^{2} + \dfrac{1}{x^{2}}$.

If $x - \dfrac{1}{x} = 4$, find $x^{2} + \dfrac{1}{x^{2}}$.
  1. $18$
  2. $14$
  3. $16$
  4. $20$

Solution

$\left(x - \dfrac{1}{x}\right)^{2} = x^{2} - 2 + \dfrac{1}{x^{2}} \Rightarrow x^{2} + \dfrac{1}{x^{2}} = 16 + 2 = 18$.

Asked in: IMO

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