Answer:
a = 11
b = 6
Step-by-step explanation:
[tex] \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a - b \sqrt{3} \\ \\ \frac{(5 + 2 \sqrt{3} )}{(7 + 4 \sqrt{3}) } \times \frac{(7 - 4 \sqrt{3})}{(7 - 4 \sqrt{3})} = a - b \sqrt{3} \\ \\ \frac{(5 + 2 \sqrt{3)} (7 - 4 \sqrt{3} )}{ {(7)}^{2} - {(4 \sqrt{3} )}^{2} } a - b \sqrt{3} \\ \\ \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24 }{49 - 48} = a - b \sqrt{3} \\ \\ \frac{11 - 6 \sqrt{3}}{1} = a - b \sqrt{3} \\ \\ 11 - 6 \sqrt{3} = a - b \sqrt{3} \\ \\ equating \: like \: terms \: on \: both \: \\ sides, \: we \: find : \\ \\ a = 11 \\ \\ b = 6[/tex]
