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Mathematics

Solve for x : 1+x+x21−x+x2=4963(1+x1−x)\dfrac{1 + x + x^2}{1 - x + x^2} = \dfrac{49}{63}\Big(\dfrac{1 + x}{1 - x}\Big).

Ratio Proportion

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Answer

Given,

⇒1+x+x21−x+x2=4963(1+x1−x)⇒(1−x)(1+x+x2)(1+x)(1−x+x2)=4963⇒1−x31+x3=4963\Rightarrow \dfrac{1 + x + x^2}{1 - x + x^2} = \dfrac{49}{63}\Big(\dfrac{1 + x}{1 - x}\Big) \\[1em] \Rightarrow \dfrac{(1 - x)(1 + x + x^2)}{(1 + x)(1 - x + x^2)} = \dfrac{49}{63} \\[1em] \Rightarrow \dfrac{1 - x^3}{1 + x^3} = \dfrac{49}{63}

Applying componendo and dividendo, we get :

⇒1−x3+1+x31−x3−(1+x3)=49+6349−63⇒21−1−x3−x3=112−14⇒−22x3=−11214⇒1x3=8⇒x3=18⇒x3=(12)3⇒x=12.\Rightarrow \dfrac{1 - x^3 + 1 + x^3}{1 - x^3 - (1 + x^3)} = \dfrac{49 + 63}{49 - 63} \\[1em] \Rightarrow \dfrac{2}{1 - 1 - x^3 - x^3} = \dfrac{112}{-14} \\[1em] \Rightarrow -\dfrac{2}{2x^3} = -\dfrac{112}{14} \\[1em] \Rightarrow \dfrac{1}{x^3} = 8 \\[1em] \Rightarrow x^3 = \dfrac{1}{8} \\[1em] \Rightarrow x^3 = \Big(\dfrac{1}{2}\Big)^3 \\[1em] \Rightarrow x = \dfrac{1}{2}.

Hence, x = 12\dfrac{1}{2}.

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