Question

root 2 + root 3 upon root 18 - root 12 equal a minus b root 6​

Answer 1

Answer:

$\frac{ \sqrt{2 } + \sqrt{3} }{ \sqrt{18} - \sqrt{12} } = a - b \sqrt{6}$

$\frac{ \sqrt{2 } + \sqrt{3} }{ \sqrt{3 \times 3 \times 2} - \sqrt{2 \times 2 \times 3} } = a - b \sqrt{6}$

$\frac{ \sqrt{2 } + \sqrt{3} }{ 3\sqrt{2} - 2\sqrt{3} } = a - b \sqrt{6}$

Rationalize

$\frac{ \sqrt{2 } + \sqrt{3} }{ 3\sqrt{2} - 2\sqrt{3} } \times \frac{3 \sqrt{2} + 2 \sqrt{3} }{3 \sqrt{2 } + 2 \sqrt{3} } = a - b \sqrt{6}$

$\frac{ \sqrt{2 } + \sqrt{3} }{ 18 - 12 } \times \frac{3 \sqrt{2} + 2 \sqrt{3} }{1 } = a - b \sqrt{6}$

$({ \sqrt{2 } + \sqrt{3} }) \times ( {3 \sqrt{2} + 2 \sqrt{3} })= 6a - b \: 6 \sqrt{6}$

$12 + 2 \sqrt{6} + 3 \sqrt{6} + 6 = 6a - 6b \sqrt{6}$

$18 + 5 \sqrt{6} = 6a - b \: 6 \sqrt{6}$

Compare both side,

$a = \frac{18}{6} = 3$

And,

$5 \sqrt{6} = - b \: 6 \sqrt{6} \\ b = - \frac{5}{6}$