Question

In the equation $(\sqrt{3} - \sqrt{2})Ax^{2} + \frac{Bx}{\sqrt{2} + \sqrt{3}}$+ C = 0 where A = (49 + $20\sqrt{6}$)^1/4, B is the sum of the infinite G.P. $8\sqrt{3} + 8\sqrt{2} + \frac{16}{\sqrt{3}}$+ ...... and the difference of the roots is $(6\sqrt{6})^{\log(10 - 2\log\sqrt{5} + \log\sqrt{\log 18 + \log 472}}$ where the base of the logrithem is 6. Then C = 2ⁿ where n =

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

Answer: (C)

7

Answer 2

A = $\sqrt{2} + \sqrt{3}$, B = $\frac{24}{\sqrt{3} - \sqrt{2}}$

x² + 24x + c = 0

a – b = 8, a + b = – 24 ® ab = 2⁷