MCQ
If $\frac{2+\sqrt{3}}{2-\sqrt{3}}=\text{a}+\text{b}\sqrt{3},$ then,
- ✓$a = 7$ and $b = 4$
- B$a = -7$ and $b = -4$
- C$a = -7$ and $b = 4$
- D$a = 7$ and $b = -4$
$\frac{2+\sqrt{3}}{2-\sqrt{3}}$
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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