Answer:
a = 11, b = 6
Step-by-step explanation:
Given
[tex]\frac{5+2\sqrt{3} }{7+4\sqrt{3} }[/tex]
Rationalise the denominator by multiplying numerator/ denominator by the conjugate of the denominator.
The conjugate of 7 + 4[tex]\sqrt{3}[/tex] is 7 - 4[tex]\sqrt{3}[/tex]
= [tex]\frac{(5+2\sqrt{3})(7-4\sqrt{3}) }{(7+4\sqrt{3})(7-4\sqrt{3} }[/tex] ← expand numerator/ denominator using FOIL
= [tex]\frac{35-20\sqrt{3}+14\sqrt{3}-24 }{49-28\sqrt{3}+28\sqrt{3}-48 }[/tex]
= [tex]\frac{11-6\sqrt{3} }{1}[/tex]
= 11 - 6[tex]\sqrt{3}[/tex]
with a = 11 and b = 6
