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Mathematics

Solve : 1+x+x21−x+x2=62(1+x)63(1−x).\dfrac{1 + x + x^2}{1 - x + x^2} = \dfrac{62(1 + x)}{63(1 - x)}.

Ratio Proportion

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Answer

Given,

1+x+x21−x+x2=62(1+x)63(1−x)\dfrac{1 + x + x^2}{1 - x + x^2} = \dfrac{62(1 + x)}{63(1 - x)}

⇒(1+x+x2)(1−x)(1−x+x2)(1+x)=6263⇒1+x+x2−x−x2−x31−x+x2+x−x2+x3=6263⇒1−x+x−x2+x2−x31+x−x−x2+x2+x3=6263⇒1−x31+x3=6263\Rightarrow \dfrac{(1 + x + x^2)(1 - x)}{(1 - x + x^2)(1 + x)} = \dfrac{62}{63} \\[1em] \Rightarrow \dfrac{1 + x + x^2 -x -x^2 - x^3}{1 - x + x^2 + x - x^2 + x^3} = \dfrac{62}{63} \\[1em] \Rightarrow \dfrac{1 - \cancel{x} + \cancel{x} - \cancel{x^2} + \cancel{x^2} - x^3}{1 + \cancel{x} - \cancel{x} - \cancel{x^2} + \cancel{x^2} + x^3} = \dfrac{62}{63} \\[1em] \Rightarrow \dfrac{1 - x^3}{1 + x^3} = \dfrac{62}{63}

Again applying componendo and dividendo,

⇒1−x3+1+x31−x3−1−x3=62+6362−63⇒2−2x3=125−1⇒−1x3=−125⇒x3=1125⇒x=11253⇒x=15.\Rightarrow \dfrac{1 - x^3 + 1 + x^3}{1 - x^3 - 1 -x^3} = \dfrac{62 + 63}{62 - 63} \\[1em] \Rightarrow \dfrac{2}{-2x^3} = \dfrac{125}{-1} \\[1em] \Rightarrow -\dfrac{1}{x^3} = -125 \\[1em] \Rightarrow x^3 = \dfrac{1}{125} \\[1em] \Rightarrow x = \dfrac{1}{\sqrt[3]{125}} \\[1em] \Rightarrow x = \dfrac{1}{5}.

Hence, the required value is 15.\dfrac{1}{5}.

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