KnowledgeBoat Logo
|

Mathematics

Find x, if 16(a−xa+x)3=a+xa−x16\Big(\dfrac{a - x}{a + x}\Big)^3 = \dfrac{a + x}{a - x}

Ratio Proportion

18 Likes

Answer

Given,

⇒16(a−xa+x)3=a+xa−x⇒16=(a+x)3.(a+x)(a−x)3.(a−x)⇒(a+x)4(a−x)4=16⇒(a+x)4(a−x)4=24⇒a+xa−x=±2\Rightarrow 16\Big(\dfrac{a - x}{a + x}\Big)^3 = \dfrac{a + x}{a - x} \\[1em] \Rightarrow 16 = \dfrac{(a + x)^3.(a + x)}{(a - x)^3.(a - x)} \\[1em] \Rightarrow \dfrac{(a + x)^4}{(a - x)^4} = 16 \\[1em] \Rightarrow \dfrac{(a + x)^4}{(a - x)^4} = 2^4 \\[1em] \Rightarrow \dfrac{a + x}{a - x} = \pm 2

In first case, let a+xa−x=2\dfrac{a + x}{a - x} = 2

Applying componendo and dividendo we get,

⇒a+x+a−xa+x−(a−x)=2+12−1⇒2a2x=31⇒ax=3⇒x=a3.\Rightarrow \dfrac{a + x + a - x}{a + x - (a - x)} = \dfrac{2 + 1}{2 - 1} \\[1em] \Rightarrow \dfrac{2a}{2x} = \dfrac{3}{1} \\[1em] \Rightarrow \dfrac{a}{x} = 3 \\[1em] \Rightarrow x = \dfrac{a}{3}.

In second case, let a+xa−x=−2\dfrac{a + x}{a - x} = -2

Applying componendo and dividendo we get,

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

Hence, x = 3a or a3\dfrac{a}{3}.

Answered By

10 Likes

Related Questions