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Mathematics

Solve for x : 16(a−xa+x)3=a+xa−x.16\Big(\dfrac{a - x}{a + x}\Big)^3 =\dfrac{a + x}{a - x}.

Ratio Proportion

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Answer

Given,

16(a−xa+x)3=a+xa−x.16\Big(\dfrac{a - x}{a + x}\Big)^3 =\dfrac{a + x}{a - x}.

⇒(a−x)3(a+x)3×(a−x)(a+x)=116⇒(a−xa+x)4=(12)4 or (−12)4⇒a−xa+x=12 or −12\Rightarrow \dfrac{(a - x)^3}{(a + x)^3} \times \dfrac{(a - x)}{(a + x)} = \dfrac{1}{16} \\[1em] \Rightarrow \Big(\dfrac{a - x}{a + x}\Big)^4 = \Big(\dfrac{1}{2}\Big)^4 \text{ or } \Big(-\dfrac{1}{2}\Big)^4 \\[1em] \Rightarrow \dfrac{a - x}{a + x} = \dfrac{1}{2} \text{ or } -\dfrac{1}{2}

First Solving,

a−xa+x=12\dfrac{a - x}{a + x} = \dfrac{1}{2}

By componendo and dividendo,

⇒a−x+a+xa−x−a−x=1+21−2⇒−2a2x=−3⇒ax=3⇒x=a3.\Rightarrow \dfrac{a - x + a + x}{a - x - a - x} = \dfrac{1 + 2}{1 - 2} \\[1em] \Rightarrow -\dfrac{2a}{2x} = -3 \\[1em] \Rightarrow \dfrac{a}{x} = 3 \\[1em] \Rightarrow x = \dfrac{a}{3}.

Now Solving,

a−xa+x=−12\dfrac{a - x}{a + x} = -\dfrac{1}{2}

By componendo and dividendo,

⇒a−x+a+xa−x−a−x=1−21+2⇒−2a2x=−13⇒ax=13⇒x=3a.\Rightarrow \dfrac{a - x + a + x}{a - x - a - x} = \dfrac{1 - 2}{1 + 2}\\[1em] \Rightarrow -\dfrac{2a}{2x} = -\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{a}{x} = \dfrac{1}{3} \\[1em] \Rightarrow x = 3a. \\[1em]

Hence, the value of x is a3\dfrac{a}{3} and 3a.

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