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If 2a+1+2a−12a+1−2a−1\dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}} = x, prove that x2 - 4ax + 1 = 0.

Ratio Proportion

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Answer

Given,

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

Applying componendo and dividendo, we get :

⇒2a+1+2a−1+2a+1−2a−12a+1+2a−1−(2a+1−2a−1)=x+1x−1⇒22a+122a−1=x+1x−1⇒2a+12a−1=x+1x−1\Rightarrow \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1} + \sqrt{2a + 1} - \sqrt{2a - 1}}{\sqrt{2a + 1} + \sqrt{2a - 1} - (\sqrt{2a+ 1} - \sqrt{2a - 1})} = \dfrac{x + 1}{x - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{2a + 1}}{2\sqrt{2a - 1}} = \dfrac{x + 1}{x - 1} \\[1em] \Rightarrow \dfrac{\sqrt{2a + 1}}{\sqrt{2a - 1}} = \dfrac{x + 1}{x - 1}

Squaring both sides we get,

⇒2a+12a−1=(x+1)2(x−1)2\Rightarrow \dfrac{2a + 1}{2a - 1} = \dfrac{(x + 1)^2}{(x - 1)^2}

Applying componendo and dividendo we get :

⇒2a+1+2a−12a+1−(2a−1)=(x+1)2+(x−1)2(x+1)2−(x−1)2⇒4a2=x2+1+2x+x2+1−2xx2+1+2x−(x2+1−2x)⇒2a=2(x2+1)4x⇒8ax=2(x2+1)⇒4ax=x2+1⇒x2−4ax+1=0.\Rightarrow \dfrac{2a + 1 + 2a - 1}{2a + 1 - (2a - 1)} = \dfrac{(x + 1)^2 + (x - 1)^2}{(x + 1)^2 - (x - 1)^2} \\[1em] \Rightarrow \dfrac{4a}{2} = \dfrac{x^2 + 1 + 2x + x^2 + 1 - 2x}{x^2 + 1 + 2x - (x^2 + 1 - 2x)} \\[1em] \Rightarrow 2a = \dfrac{2(x^2 + 1)}{4x} \\[1em] \Rightarrow 8ax = 2(x^2 + 1) \\[1em] \Rightarrow 4ax = x^2 + 1 \\[1em] \Rightarrow x^2 - 4ax + 1 = 0.

Hence, proved that x2 - 4ax + 1 = 0.

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