Question

7+3√5 / 7-3√5 = a+b√5

Answer 1

$\frac{7+3 \sqrt{5} }{7-3 \sqrt{5} } = a+b \sqrt{5}$

By Rationalization,

$\frac{(7+3 \sqrt{5})*(7+3 \sqrt{5}) }{(7-3 \sqrt{5})*(7+3 \sqrt{5}) }$


Now,
_________________________________
On nominator, 

by $(a+b)^{2} = a^{2} + b^{2} + 2ab$

On denominator,

By $a^{2} - b^{2} = (a+b) (a-b)$
______________________________

Then,

$\frac{(7+3 \sqrt{5})^{2} }{(7)^{2}- (3 \sqrt{5})^{2} }$

$\frac{49+ 45 + 6 \sqrt{5} }{49 - 45}$

= $\frac{94+6 \sqrt{5} }{4}$

Now,


a =  $\frac{94}{4}$ = $\frac{47}{2}$
b = $\frac{6 \sqrt{5} }{4} = \frac{3 \sqrt{5} }{2}$


i hope this will help you


-by ABHAY

Answer 2

Answer:

Hope this helped you.

Thank you.