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

Ratio Proportion

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Answer

Given,

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

Applying componendo and dividendo on both sides we get :

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

Squaring both sides we get :

⇒(x+1)2(x−1)2=2a+12a−1⇒x2+1+2xx2+1−2x=2a+12a−1\Rightarrow \dfrac{(x + 1)^2}{(x - 1)^2} = \dfrac{2a + 1}{2a - 1} \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{2a + 1}{2a - 1}

Applying componendo and dividendo again :

⇒x2+1+2x+x2+1−2xx2+1+2x−(x2+1−2x)=2a+1+2a−12a+1−(2a−1)⇒2x2+2x2−x2+1−1+2x−(−2x)=4a2a−2a+1−(−1)⇒2x2+24x=4a2⇒2(x2+1)4x=2a⇒x2+12x=2a⇒x2+1=4ax⇒x2−4ax+1=0.\Rightarrow \dfrac{x^2 + 1 + 2x + x^2 + 1 - 2x}{x^2 + 1 + 2x - (x^2 + 1 - 2x)} = \dfrac{2a + 1 + 2a - 1}{2a + 1 - (2a - 1)} \\[1em] \Rightarrow \dfrac{2x^2 + 2}{x^2 - x^2 + 1 - 1 + 2x - (-2x)} = \dfrac{4a}{2a - 2a + 1 - (-1)} \\[1em] \Rightarrow \dfrac{2x^2 + 2}{4x} = \dfrac{4a}{2} \\[1em] \Rightarrow \dfrac{2(x^2 + 1)}{4x} = 2a \\[1em] \Rightarrow \dfrac{x^2 + 1}{2x} = 2a \\[1em] \Rightarrow x^2 + 1 = 4ax \\[1em] \Rightarrow x^2 - 4ax + 1 = 0.

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

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