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Mathematics

If x = a+1+a−1a+1−a−1,\dfrac{\sqrt{a + 1} + \sqrt{a - 1}}{\sqrt{a + 1} - \sqrt{a - 1}}, using properties of proportion, show that

x2 - 2ax + 1 = 0.

Ratio Proportion

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Answer

Given,

x1=a+1+a−1a+1−a−1.\dfrac{x}{1} = \dfrac{\sqrt{a + 1} + \sqrt{a - 1}}{\sqrt{a + 1} - \sqrt{a - 1}}.

Applying componendo and dividendo,

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

Squaring both sides we get,

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

Again applying componendo and dividendo,

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

Hence proved that, x2 - 2ax + 1 = 0.

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