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

Applying componendo and dividendo, we get :

⇒x+1x−1=2a+1+2a−1+2a+1−2a−12a+1+2a−1−(2a+1−2a−1)⇒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{\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}}

Squaring both sides, we get :

⇒(x+1)2(x−1)2=2a+12a−1⇒x2+1+2xx2+1−2x=2a+12a−1⇒(x2+1+2x)(2a−1)=(x2+1−2x)(2a+1)⇒2ax2−x2+2a−1+4ax−2x=2ax2+x2+2a+1−4ax−2x⇒2ax2−2ax2+x2+x2+2a−2a+1+1−4ax−4ax−2x+2x=0⇒2x2−8ax+2=0⇒2(x2−4ax+1)=0⇒x2−4ax+1=0.\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] \Rightarrow (x^2 + 1 + 2x)(2a - 1) = (x^2 + 1 - 2x)(2a + 1) \\[1em] \Rightarrow 2ax^2 - x^2 + 2a - 1 + 4ax - 2x = 2ax^2 + x^2 + 2a + 1 - 4ax - 2x \\[1em] \Rightarrow 2ax^2 - 2ax^2 + x^2 + x^2 + 2a - 2a + 1 + 1 - 4ax - 4ax - 2x + 2x = 0 \\[1em] \Rightarrow 2x^2 - 8ax + 2 = 0 \\[1em] \Rightarrow 2(x^2 - 4ax + 1) = 0 \\[1em] \Rightarrow x^2 - 4ax + 1 = 0.

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

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