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Mathematics

If x=m+n+m−nm+n−m−nx = \dfrac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}}, express n in terms of x and m.

Ratio Proportion

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Answer

Given,

x=m+n+m−nm+n−m−nx = \dfrac{\sqrt{m + n} + \sqrt{m - n}}{\sqrt{m + n} - \sqrt{m - n}}

Applying componendo and dividendo,

⇒x+1x−1=m+n+m−n+m+n−m−nm+n+m−n−(m+n−m−n)⇒x+1x−1=2m+n2m−n⇒x+1x−1=m+nm−n⇒m+nm−n=(x+1)2(x−1)2⇒m+nm−n=x2+1+2xx2+1−2x\Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{m + n} + \sqrt{m - n} + \sqrt{m + n} - \sqrt{m - n}}{\sqrt{m + n} + \sqrt{m - n} - (\sqrt{m + n} - \sqrt{m - n})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{m + n}}{2\sqrt{m - n}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{m + n}}{\sqrt{m - n}} \\[1em] \Rightarrow \dfrac{m + n}{m - n} = \dfrac{(x + 1)^2}{(x - 1)^2} \\[1em] \Rightarrow \dfrac{m + n}{m - n} = \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} \\[1em]

Applying componendo and dividendo,

⇒m+n+m−nm+n−(m−n)=x2+1+2x+x2+1−2xx2+1+2x−(x2+1−2x)⇒2m2n=2(x2+1)4x⇒mn=x2+12x⇒n=2mxx2+1.\Rightarrow \dfrac{m + n + m - n}{m + n - (m - n)} = \dfrac{x^2 + 1 + 2x + x^2 + 1 - 2x}{x^2 + 1 + 2x - (x^2 + 1 - 2x)} \\[1em] \Rightarrow \dfrac{2m}{2n} = \dfrac{2(x^2 + 1)}{4x} \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{x^2 + 1}{2x} \\[1em] \Rightarrow n = \dfrac{2mx}{x^2 + 1}.

Hence, n = 2mxx2+1.\dfrac{2mx}{x^2 + 1}.

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