Question

Please answer this question

Answer 1

ANSWER.

• a = 16.
• b = -1.

SOLUTION.

Given –

$\tt \implies \dfrac{2 + \sqrt{3} }{2-\sqrt{3}} + \dfrac{2 - \sqrt{3} }{2 + \sqrt{3} } + \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1} = a + b \sqrt{3}$

• To solve this, we have to rationalise the denominator of the fractions.

$\tt \dfrac{2 + \sqrt{3} }{2 - \sqrt{3} }$

$\tt = \dfrac{(2 + \sqrt{3})^{2} }{(2 - \sqrt{3})(2 + \sqrt{3})}$

$\tt = \dfrac{4+ 3 + 4 \sqrt{3} }{4 - 3}$

$\tt = 7 + 4 \sqrt{3}$

Similarly,

$\tt \dfrac{2 - \sqrt{3} }{2 + \sqrt{3} }$

$\tt = \dfrac{(2 - \sqrt{3})^{2} }{(2 + \sqrt{3})(2 - \sqrt{3}) }$

$\tt = \dfrac{4 + 3 - 4 \sqrt{3} }{4 - 3}$

$\tt = 7 - 4 \sqrt{3}$

Also,

$\tt \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1}$

$\tt = \dfrac{( \sqrt{3} - 1)^{2} }{( \sqrt{3} + 1)( \sqrt{3} - 1) }$

$\tt = \dfrac{3 + 1 - 2 \sqrt{3} }{3 - 1}$

$\tt = \dfrac{4- 2 \sqrt{3} }{2}$

$\tt =2 - \sqrt{3}$

Therefore,

$\tt \implies \dfrac{2 + \sqrt{3} }{2-\sqrt{3}} + \dfrac{2 - \sqrt{3} }{2 + \sqrt{3} } + \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1} = a + b \sqrt{3}$

$\tt \implies7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} + 2 - \sqrt{3} = a + b \sqrt{3}$

$\tt \implies16 - \sqrt{3} = a + b \sqrt{3}$

Comparing both sides, we get,

$\begin{cases} \tt a = 16 \\ \tt b = - 1 \end{cases}$

Which is our required answer.