Question
If $\frac{2+\sqrt{3}}{2-\sqrt{3}}=\text{a}+\text{b}\sqrt{3},$ then,

Answer

  1. a = 7 and b = 4
    Solution:
    $\frac{2+\sqrt{3}}{2-\sqrt{3}}$
    Multiplying numerator and denominator by $2\div\sqrt{3}$
    So, $\frac{(2+\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}$
    $=\frac{4+3+4\sqrt{3}}{4-3},$
    $=7\div4\sqrt{3}$
    Now equating $7\div4\sqrt{3}$ and $\text{a}\div\text{b}\sqrt{3}$
    we get,
    a = 7 and b = 4

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