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Mathematics

Solve for x, using the properties of proportionality :

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

Applying componendo and dividendo, we get :

⇒1+x+x2+1−x+x21+x+x2−(1−x+x2)=62(1+x)+63(1−x)62(1+x)−63(1−x)⇒2+2x21−1+x+x+x2−x2=62+62x+63−63x62+62x−63+63x⇒2(1+x2)2x=125−x125x−1⇒1+x2x=125−x125x−1⇒(125x−1)(1+x2)=x(125−x)⇒125x+125x3−1−x2=125x−x2⇒125x−125x−x2+x2+125x3−1=0⇒125x3−1=0⇒125x3=1⇒x3=1125⇒x3=(15)3⇒x=15.\Rightarrow \dfrac{1 + x + x^2 + 1 - x + x^2}{1 + x + x^2 - (1 - x + x^2)} = \dfrac{62(1 + x) + 63(1 - x)}{62(1 + x) - 63(1 - x)} \\[1em] \Rightarrow \dfrac{2 + 2x^2}{1 - 1 + x + x + x^2- x^2} = \dfrac{62 + 62x + 63 - 63x}{62 + 62x - 63 + 63x} \\[1em] \Rightarrow \dfrac{2(1 + x^2)}{2x} = \dfrac{125 - x}{125x - 1} \\[1em] \Rightarrow \dfrac{1 + x^2}{x} = \dfrac{125 - x}{125x - 1} \\[1em] \Rightarrow (125x - 1)(1 + x^2) = x(125 - x)\\[1em] \Rightarrow 125x + 125x^3 - 1 - x^2 = 125x - x^2 \\[1em] \Rightarrow 125x - 125x - x^2 + x^2 + 125x^3 - 1 = 0 \\[1em] \Rightarrow 125x^3 - 1 = 0 \\[1em] \Rightarrow 125x^3 = 1 \\[1em] \Rightarrow x^3 = \dfrac{1}{125} \\[1em] \Rightarrow x^3 = \Big(\dfrac{1}{5}\Big)^3 \\[1em] \Rightarrow x = \dfrac{1}{5}.

Hence, x = 15\dfrac{1}{5}.

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