Question

Find a and b from the above question. Please give step by step explanation also.​

Answer 1

$so \: the \: values \: are \: \boxed{a = 0} \: and \: \boxed{b = 1}$

Step-by-step explanation:

$\frac{7 + \sqrt{5} }{7 - \sqrt{5} } - \frac{7 - \sqrt{5} }{7 + \sqrt{5} } = a + b\frac{7}{11}\sqrt{5}$

$\frac{(7 + \sqrt{5} )(7 + \sqrt{5} )}{(7 - \sqrt{5})(7 + \sqrt{5} ) } - \frac{(7 - \sqrt{5} )(7 - \sqrt{5} )}{(7 + \sqrt{5} )(7 - \sqrt{5} )} = a + b\frac{7}{11} \sqrt{5}$

$\frac{(7 + \sqrt{5} )^{2} }{ {(7)}^{2} - {( \sqrt{5} )}^{2} } - \frac{(7 - \sqrt{5} )^{2} }{ {(7)}^{2} - {( \sqrt{5} )}^{2} } = a + b\frac{7}{11} \sqrt{5}$

$\frac{(7)^{2} + 2 \times 7 \times \sqrt{5} + {( \sqrt{5} )}^{2} }{49 - 5} - \frac{(7)^{2} - 2 \times 7 \times \sqrt{5} }{49 - 5} = a + b \frac{7}{11} \sqrt{5}$

$\frac{49 + 5 + 14 \sqrt{5} }{44} - \frac{49 + 5 - 14 \sqrt{5}}{44} = a + b \frac{7}{11} \sqrt{5}$

$\frac{54 + 14 \sqrt{5} - (54 - 14 \sqrt{5} ) }{44} = a + b \frac{7}{11} \sqrt{5}$

$\frac{54 + 14 \sqrt{5} - 54 + 14 \sqrt{5} }{44} = a + b \frac{7}{11} \sqrt{5}$

$\frac{28 \sqrt{5} }{44} = a + b \frac{7}{11} \sqrt{5}$

$\frac{7}{11} \sqrt{5} = a + b \frac{7}{11} \sqrt{5}$

$by \: comparision \: of \: both \: sides \: we \: get$

$a = 0 \: and \: b = 1$

Answer 2

Answer:

$\frac{(7 + \sqrt{5} )(7 + \sqrt{5} )}{(7 - \sqrt{5})(7 + \sqrt{5} ) } - \frac{(7 - \sqrt{5} )(7 - \sqrt{5} )}{(7 + \sqrt{5} )(7 - \sqrt{5} )} = a + b\frac{7}{11} \sqrt{5} \\ \\ \frac{(7 + \sqrt{5} )^{2} }{ {(7)}^{2} - {( \sqrt{5} )}^{2} } - \frac{(7 - \sqrt{5} )^{2} }{ {(7)}^{2} - {( \sqrt{5} )}^{2} } = a + b\frac{7}{11} \sqrt{5} \\ \\ \frac{(7)^{2} + 2 \times 7 \times \sqrt{5} + {( \sqrt{5} )}^{2} }{49 - 5} - \frac{(7)^{2} - 2 \times 7 \times \sqrt{5} }{49 - 5} = a + b \frac{7}{11} \sqrt{5} \\ \\ \frac{49 + 5 + 14\sqrt{5} }{44} - \frac{49 + 5 - 14 \sqrt{5}}{44} = a + b \frac{7}{11} \sqrt{5} \\ \\ \frac{54 + 14 \sqrt{5} - (54 - 14 \sqrt{5} ) }{44} = a + b \frac{7}{11} \sqrt{5} \\ \\ </p><p>\frac{54 + 14 \sqrt{5} - 54 + 14 \sqrt{5} }{44} = a + b \frac{7}{11} \sqrt{5} \\ \frac{28 \sqrt{5} }{44} = a + b \frac{7}{11} \sqrt{5} \\ \\ \frac{7}{11} \sqrt{5} = a + b \\ \\ by \: comparision \: of \: both \: sides \: we \: \\ \\ a = 0 \: and \: b = 1a=0andb=1$