Question

$(2 \sqrt{3} + 3 \sqrt{2} ) \div 2 \sqrt{3} - 3 \sqrt{2} = a - b \sqrt{6}$
Please find a and b​

Answer 1

Answer:

$Value \: of \: a = - 5 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: b = 2$

Step-by-step explanation:

$Since,$

$\frac{(2 \sqrt{3} + 3 \sqrt{2}) }{(2 \sqrt{3} - 3 \sqrt{2}) } = \frac{2 \sqrt{3} + 3 \sqrt{2} }{2 \sqrt{3} - 3 \sqrt{2} } \times \frac{2 \sqrt{3} + 3 \sqrt{2} }{2 \sqrt{3} + 3 \sqrt{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{4 \times 3 + 9 \times 2 + 2 \times 2 \sqrt{3} \times 3 \sqrt{2} }{(2 \sqrt{3}) {}^{2} -( 3 \sqrt{2} ) {}^{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{12 + 18 + 12 \sqrt{6} }{12 - 18} = \frac{30 + 12 \sqrt{6} }{ - 6}$

$that \: is \: = > - 5 - 2 \sqrt{6}$

$Given \: expression...$

$\frac{2 \sqrt{3} + 3 \sqrt{2} }{2 \sqrt{3} - 3 \sqrt{2} } = a - b \sqrt{6} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > (- 5) - ( - 2 \sqrt{6} ) \\ that \: is = > ( - 5) + 2 \sqrt{6}$

$Therefore \: a = - 5 \: and \: b = 2$.

Hope this proves helpful -ᄒᴥᄒ-