Question

If 3+√7/3-√7=2a+3b√7 find a and b

Answer 1

Answer:

a=4 and b=1

Steps

Given

$\frac{3 + \sqrt{7} }{3 - \sqrt{7} } = 2a + 3b \sqrt{7}$

To find a and b

Solve LHS

$\frac{3 + \sqrt{7} }{3 - \sqrt{7} } \\ multipy \: numerator \: and \: denomintor \: by \: 3 + \sqrt{7} \\ = \frac{3 + \sqrt{7} }{3 - \sqrt{7} } \times \frac{3 + \sqrt{7} }{3 + \sqrt{7} } \\ = \frac{9 + 3 \sqrt{7} + 3 \sqrt{7} + 7 }{ {3}^{2} - { (\sqrt{7}) }^{2} } \: ({{as \: {a}^{2} - {b}^{2} = ( + b)(a - b)}}) \\ \\ = \frac{16 + 6 \sqrt{7}}{9 - 7} \\ = \frac{16 + 6 \sqrt{7}}{2} \\ = 8 + 3 \sqrt{7}$

Now,

$2a + 3b \sqrt{7} = 8 + 3 \sqrt{7} \\ 2a + 3b \sqrt{7} =( 2 \times 4) + (3 \times 1 )\sqrt{7}$

Hence, a=4 and b=1