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If x = 12mnm+n\dfrac{12mn}{m + n}, find the value of :

x+6mx−6m+x+6nx−6n\dfrac{x + 6m}{x - 6m} + \dfrac{x + 6n}{x - 6n}.

Quadratic Equations

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Answer

Substituting value of x in x+6mx−6m+x+6nx−6n\dfrac{x + 6m}{x - 6m} + \dfrac{x + 6n}{x - 6n}, we get :

⇒12mnm+n+6m12mnm+n−6m+12mnm+n+6n12mnm+n−6n⇒12mn+6m(m+n)m+n12mn−6m(m+n)m+n+12mn+6n(m+n)m+n12mn−6n(m+n)m+n⇒12mn+6m(m+n)12mn−6m(m+n)+12mn+6n(m+n)12mn−6n(m+n)⇒12mn+6mn+6m212mn−6mn−6m2+12mn+6mn+6n212mn−6mn−6n2⇒18mn+6m26mn−6m2+18mn+6n26mn−6n2⇒6m(3n+m)6m(n−m)+6n(3m+n)6n(m−n)⇒3n+mn−m+3m+nm−n⇒3n+mn−m+3m+n−(n−m)⇒3n+mn−m−3m+nn−m⇒3n+m−(3m+n)n−m⇒3n+m−3m−nn−m⇒2n−2mn−m⇒2(n−m)(n−m)⇒2.\Rightarrow \dfrac{\dfrac{12mn}{m + n} + 6m}{\dfrac{12mn}{m + n} - 6m} + \dfrac{\dfrac{12mn}{m + n} + 6n}{\dfrac{12mn}{m + n} - 6n} \\[1em] \Rightarrow \dfrac{\dfrac{12mn + 6m(m + n)}{m + n}}{\dfrac{12mn - 6m(m + n)}{m + n}} + \dfrac{\dfrac{12mn + 6n(m + n)}{m + n}}{\dfrac{12mn - 6n(m + n)}{m + n}} \\[1em] \Rightarrow \dfrac{12mn + 6m(m + n)}{12mn - 6m(m + n)} + \dfrac{12mn + 6n(m + n)}{12mn - 6n(m + n)} \\[1em] \Rightarrow \dfrac{12mn + 6mn + 6m^2}{12mn - 6mn - 6m^2} + \dfrac{12mn + 6mn + 6n^2}{12mn - 6mn - 6n^2} \\[1em] \Rightarrow \dfrac{18mn + 6m^2}{6mn - 6m^2} + \dfrac{18mn + 6n^2}{6mn - 6n^2} \\[1em] \Rightarrow \dfrac{6m(3n + m)}{6m(n - m)} + \dfrac{6n(3m + n)}{6n(m - n)} \\[1em] \Rightarrow \dfrac{3n + m}{n - m} + \dfrac{3m + n}{m - n} \\[1em] \Rightarrow \dfrac{3n + m}{n - m} + \dfrac{3m + n}{-(n - m)} \\[1em] \Rightarrow \dfrac{3n + m}{n - m} - \dfrac{3m + n}{n - m} \\[1em] \Rightarrow \dfrac{3n + m - (3m + n)}{n - m} \\[1em] \Rightarrow \dfrac{3n + m - 3m - n}{n - m} \\[1em] \Rightarrow \dfrac{2n - 2m}{n - m} \\[1em] \Rightarrow \dfrac{2(n - m)}{(n - m)} \\[1em] \Rightarrow 2.

Hence, x+6mx−6m+x+6nx−6n=2\dfrac{x + 6m}{x - 6m} + \dfrac{x + 6n}{x - 6n} = 2.

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