Question

Find the value of a and b
$\frac{7 + \sqrt{5} }{7 - \sqrt{5} } - \frac{7 - \sqrt{5} }{7 + \sqrt{5} } = a + \frac{7}{11} \sqrt{5b}$
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Answer 1

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Find the value of a and b

$\bf \frac{7+\sqrt{5} }{7-\sqrt{5} } - \frac{7-\sqrt5}{7+\sqrt5} = a + \frac{7\sqrt{5b}}{11}$

$\bullet\square{\frak{\orange{AnsWer}}}\square\bullet$

Solving LHS

$\bf LHS = \frac{7+\sqrt5}{7-\sqrt5} - \frac{7-\sqrt5}{7+\sqrt5} \\\\\bf (taking\: LCM)\:and\:using\:identity\\\bf (a+b)(a-b)=a^2-b^2\\\\= \bf \frac{(7+\sqrt5)^2-(7-\sqrt5)^2}{(7)^2 - (\sqrt5)^2} \\\\\bf = \frac{49+5+14\sqrt5-(49+5-14\sqrt5)}{49-5} \\\\=\bf \frac{49 + 5+14\sqrt5-49-5+14\sqrt5}{44} \\\\\bf = \frac{28\sqrt5}{44}$

$\bf = \frac{7\sqrt5}{11}$

Comparing simplified LHS with RHS

$\bf \frac{7\sqrt5}{11} =a+ \frac{7\sqrt{5b}}{11} \\\\\bf 0 + \frac{7\sqrt{5(1)}}{11} =a+ \frac{7\sqrt{5b}}{11} \\\\on\:comparison\\\\\boxed{\bf a = 0}\\\\and\\\\\boxed{\bf b = 1}$