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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}}, using properties of proportion, 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−1Squaring both sides we get :⇒(x+1)2(x−1)2=2a+12a−1⇒x2+1+2xx2+1−2x=2a+12a−1Applying componendo and dividendo on both sides we get :⇒x2+1+2x+x2+1−2xx2+1+2x−(x2+1−2x)=2a+1+2a−12a+1−(2a−1)⇒2x2+24x=4a2⇒x2+12x=2a⇒x2+1=4ax⇒x2+1−4ax=0\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] \text{Squaring both sides we get :}\\[1em] \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}\\[1em] \text{Applying componendo and dividendo on both sides we get :}\\[1em] \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}{4x} = \dfrac{4a}{2}\\[1em] \Rightarrow \dfrac{x^2 + 1}{2x} = 2a\\[1em] \Rightarrow x^2 + 1 = 4ax\\[1em] \Rightarrow x^2 + 1 - 4ax = 0

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

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