Question

$n = \sqrt{p + 2q \: } \: + \sqrt{p - 2q } \div \sqrt{p + 2q} - \sqrt{p - 2q}$
is given than prove that
$qn {}^{2} - pn + q = 0$

Answer 1

$n = \frac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} - \sqrt{p - 2q} } \\ rationalise \: \\ n = \frac{( \sqrt{p + 2q} + \sqrt{p - 2q}) {}^{2} }{p + 2q - p +2q} \\ 4nq = p + 2q + p - 2q + 2 \sqrt{ {p}^{2} - 4 {q}^{2} } \\ 4nq = 2p + 2 \sqrt{p {}^{2} - 4 {q}^{2} } \\ (2nq - p) = \sqrt{p {}^{2} - 4 {q}^{2} } \\ square \\ (4 {n}^{2} {q}^{2} + p {}^{2} - 4pnq) = {p}^{2} - 4q {}^{2} \\ 4n {}^{2} {q}^{2} - 4npq = - 4 {q}^{2} \\ {n}^{2} q - np = - q$
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