Question

plz help me to solve this !​

Answer 1

Answer:

a = 4 & b = 6

Step-by-step explanation:

$LHS = \frac{a + \sqrt{3} }{2 - \sqrt{3} } \\ On \: factorization \\ \\ = > \frac{a + \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ \\ = > \frac{(a + \sqrt{3})(2 + \sqrt{3} ) }{(2 - \sqrt{3})(2 + \sqrt{3} ) } \\ \\ = > \frac{2a + \sqrt{3} a + 2 \sqrt{3} + 3}{ {2}^{2} - { (\sqrt{3}) }^{2} } \\ \\ = > \frac{(2a + 3) + \sqrt{3} (a + 2)}{4 - 3} \\ \\ = > (2a + 3) + \sqrt{3} (a + 2) \\ \\ Now \: RHS \: = 11 + b \sqrt{3} \\ on \: comparing \: LHS \: and \: RHS \\ we \: get \\ 2a + 3 = 11 \: \: and \: \: a + 2 = b \\ = > 2a = 8 \\ = > a = 4 \\ \\ = > a + 2 = b \\ = > 4 + 2 = b \\ = > b = 6 \\ \\ Therefore \: \: a = 4 \: \: and \: \: b = 6$