Question

find the values of a and b ​

Answer 1

$\sf \rightarrow \quad \frac{7 + 3 \sqrt{5} }{ 3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{ 3 - \sqrt{5} } = a + \sqrt{5} b \\ \\ \sf \rightarrow \quad \frac{3 - \sqrt{5} (7 + 3 \sqrt{5}) - (3 + \sqrt{5} ) (7 - 3 \sqrt{5} )}{(3 + \sqrt{5} )(3 - \sqrt{5} )} = a + b \sqrt{5} \\ \\ \sf \rightarrow \quad \frac{21 + 9 \sqrt{5} - 7 \sqrt{5} - 15 - \{ 21 - 9\sqrt{5} + 7\sqrt{5} - 15 \} }{9 - 5} = a + b \sqrt{5} \\ \\ \sf \rightarrow \quad \frac{\cancel{21} + 9 \sqrt{5} - 7 \sqrt{5} - \cancel{15} - \cancel{21} + 9\sqrt{5} - 7\sqrt{5} + \cancel{15} \} }{4} = a + b \sqrt{5} \\ \\ </p><p></p><p>\sf \rightarrow \quad \frac{18\sqrt{5} - 14\sqrt{5}}{4} = a + b\sqrt{5} = a + b\sqrt{5} \\ \\</p><p></p><p>\sf \rightarrow \quad \frac{\cancel{4}\sqrt{5}}{\cancel{4}} = a + b\sqrt{5}</p><p></p><p>\\ \\</p><p></p><p>\sf \rightarrow \quad 0 + 1 \cdot \sqrt{5} = a + b \cdot \sqrt{5} \\ \\</p><p></p><p> \sf \rightarrow \quad \: a = 0 \quad , \: \: \: b = 1$

Answer 2

Answer:

in this question we have to rationalize

Step-by-step explanation:

hee is your answer in the attachments

hope this will help you

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