Question

If x = √(m+n) + √(m-n)/√(m+n) - √(m-n) ​​​​​​, where n≠0, then find the value of nx2 - 2mx + n.

Answer 1

$The \: \: answer \: \: is \: \: given \: \: below: \\ \\ Now ,\\ \\ x = \frac{ \sqrt{m + n} + \sqrt{m - n} }{ \sqrt{m + n} - \sqrt{m - n} } \\ We \: \: rationalize \: \: the \: \: denominator \: \: \\ by \: \: multiplying \: \: both \: \: the \\ numerator \: \: and \: \: denominator \\ by \: \: ( \sqrt{m + n} + \sqrt{m - n} ). \\ \\ So, \\ \\ x = \frac{( \sqrt{m + n} + \sqrt{m - n} )( \sqrt{m + n} + \sqrt{m - n} )}{( \sqrt{m + n} - \sqrt{m - n})( \sqrt{m + n} + \sqrt{m - n} )} \\ \\ = \frac{(m + n) + 2 \sqrt{(m + n)(m - n) }+ (m - n)}{(m + n) - (m - n)} \\ \\ = \frac{2m + 2 \sqrt{ {m}^{2} - {n}^{2} } }{2n} \\ \\ = \frac{m + \sqrt{ {m}^{2} - {n}^{2} } }{n} \\ \\ Or, \: \: nx = m + \sqrt{ {m}^{2} - {n}^{2} } \\ \\ Or, \: \: nx - m = \sqrt{ {m}^{2} - {n}^{2} } \\ \\ Now, \: \: squaring \: \: both \: \: sides, \\ we \: \: get \\ \\ {(nx - m)}^{2} = {( \sqrt{ {m}^{2} - {n}^{2} }) }^{2} \\ \\ Or,\: \: {n}^{2} {x}^{2} - 2mnx + {m}^{2} = {m}^{2} - {n}^{2} \\ \\ Or, \: \: {n}^{2} {x}^{2} - 2mnx = - {n}^{2} \\ \\ Or, \: \: n {x}^{2} - 2mx = - n, \\ since \: \: n \: \: is \: \: non \: \: zero \\ \\ Or, \: \: n {x}^{2} - 2mx + n = 0 \\ \\ Hence, \: \: n {x}^{2} - 2mx + n = 0 \: \: (Answer) \\ \\ RULES: \\ \\ 1. \: \: x \times x= {x}^{2} \\ \\ 2. \: \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ \\ 3. \: \: (x + y)(x - y) = {x}^{2} - {y}^{2} \\ \\ 4. \: \: {( \sqrt{x}) }^{2} = x \\ \\ Thank \: \: you \: \: for \: \: your \: \: question.$

Answer 2

here is your answer in this picture