Find the value of $$\sqrt{\frac{2 + \sqrt3}{2 - \sqrt3}}$$
Correct to three places of decimal.
Let us consider $$\frac{2 + \sqrt3}{2 - \sqrt3}$$
Rationalising the denominator by multiplying and diving with 2+$$\sqrt3$$ we get,
$$\frac{(2 + \sqrt3)\times (2 + \sqrt3) }{(2 - \sqrt3)\times (2 + \sqrt3) } = \frac {(2 + \sqrt3)^2}{4 - 3} = (2 + \sqrt3)^2$$
Now,
$$\sqrt{\frac{2 + \sqrt3}{2 - \sqrt3}} = \sqrt{(2 + \sqrt3)^2} = 2 + \sqrt3 = 2 + 1.732 = 3.732$$
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