Question

if a and b are rational numbers and
$4 + 3 \sqrt{5 } \div 4 - 3 \sqrt{5} = a + b \sqrt{5}$
then find a and b

Answer 1

$= > \frac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } = a + b \sqrt{5}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: By \: \: Rationalization ,$


$= > \frac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } \times \frac{4 + 3 \sqrt{5} }{4 + 3 \sqrt{5} } = a + b \sqrt{5} \\ \\ \\ \\ = > \frac{ {( 4 + 3 \sqrt{5}) }^{2} }{ {4}^{2} - {(3 \sqrt{5})}^{2} } = a + b \sqrt{5} \\ \\ \\ \\ = > \frac{16 + 45 + 24 \sqrt{5} }{16 - 45} = a + b \sqrt{5} \\ \\ \\ \\ = > \frac{61 + 24 \sqrt{5} }{ - 29} = a + b \sqrt{5} \\ \\ \\ \\ = > \frac{ - 61 - 24 \sqrt{5} }{29} = a + b \sqrt{5} \\ \\ \\ \\ = > \frac{ - 61}{29} - \frac{24 \sqrt{3} }{29} = a + b \sqrt{5}$





On comparing values of both sides , we get that in left hand side -24 is with root 3 and in right hand side b has root 3

So,

- 24 / 29 = b

- 61 / 29 = a