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If 16(a−xa+x)3=(a+xa−x)16\Big(\dfrac{a - x}{a + x}\Big)^3 = \Big(\dfrac{a + x}{a - x}\Big); show that : a = 3x.

Quadratic Equations

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Answer

Given,

16(a−xa+x)3=(a+xa−x)⇒16(a−xa+x)3=1a−xa+x16\Big(\dfrac{a - x}{a + x}\Big)^3 = \Big(\dfrac{a + x}{a - x}\Big) \\[1em] \Rightarrow 16\Big(\dfrac{a - x}{a + x}\Big)^3 = \dfrac{1}{\dfrac{a - x}{a + x}}

Let (a−xa+x)\Big(\dfrac{a - x}{a + x}\Big) = t.

Substituting (a−xa+x)\Big(\dfrac{a - x}{a + x}\Big) = t in given equation, we get :

⇒16t3=1t⇒16t4=1⇒t4=116⇒t4=(12)4⇒t=12⇒a−xa+x=12⇒2(a−x)=a+x⇒2a−2x=a+x⇒2a−a=x+2x⇒a=3x.\Rightarrow 16t^3 = \dfrac{1}{t} \\[1em] \Rightarrow 16t^4 = 1 \\[1em] \Rightarrow t^4 = \dfrac{1}{16} \\[1em] \Rightarrow t^4 = \Big(\dfrac{1}{2}\Big)^4 \\[1em] \Rightarrow t = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{a - x}{a + x} = \dfrac{1}{2} \\[1em] \Rightarrow 2(a - x) = a + x \\[1em] \Rightarrow 2a - 2x = a + x \\[1em] \Rightarrow 2a - a = x + 2x \\[1em] \Rightarrow a = 3x.

Hence, proved that a = 3x.

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