If $x = 2 + \sqrt{3}$, find the value of $x + \dfrac{1}{x}$.

If $x = 2 + \sqrt{3}$, find the value of $x + \dfrac{1}{x}$.
  1. $4$
  2. $2\sqrt{3}$
  3. $2$
  4. $1$

Solution

Rationalise: $\dfrac{1}{x} = \dfrac{1}{2+\sqrt{3}} \cdot \dfrac{2-\sqrt{3}}{2-\sqrt{3}} = \dfrac{2-\sqrt{3}}{4-3} = 2 - \sqrt{3}$. Sum = $(2+\sqrt{3}) + (2-\sqrt{3}) = 4$.

Asked in: IMO

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