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If x=2a+3b+2a−3b2a+3b−2a−3bx = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b}}{\sqrt{2a + 3b} - \sqrt{2a - 3b}}, prove that : 3bx2 - 4ax + 3b = 0.

Ratio Proportion

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Answer

Given,

⇒x=2a+3b+2a−3b2a+3b−2a−3b\Rightarrow x = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b}}{\sqrt{2a + 3b} - \sqrt{2a - 3b}}

Applying componendo and dividendo, we get :

⇒x+1x−1=2a+3b+2a−3b+2a+3b−2a−3b2a+3b+2a−3b−(2a+3b−2a−3b)⇒x+1x−1=2a+3b+2a−3b+2a+3b−2a−3b2a+3b+2a−3b−2a+3b+2a−3b⇒x+1x−1=22a+3b22a−3b⇒x+1x−1=2a+3b2a−3b\Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b} + \sqrt{2a + 3b} - \sqrt{2a - 3b}}{\sqrt{2a + 3b} + \sqrt{2a - 3b} - (\sqrt{2a + 3b} - \sqrt{2a - 3b})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b} + \sqrt{2a + 3b} - \sqrt{2a - 3b}}{\sqrt{2a + 3b} + \sqrt{2a - 3b} - \sqrt{2a + 3b} + \sqrt{2a - 3b}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{2a + 3b}}{2\sqrt{2a - 3b}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 3b}}{\sqrt{2a - 3b}}

Squaring both sides, we get :

⇒(x+1)2(x−1)2=2a+3b2a−3b⇒x2+1+2xx2+1−2x=2a+3b2a−3b⇒(x2+1+2x)(2a−3b)=(x2+1−2x)(2a+3b)⇒2ax2+2a+4ax−3bx2−3b−6bx=2ax2+2a−4ax+3bx2+3b−6bx⇒2ax2−2ax2+2a−2a+4ax+4ax−3bx2−3bx2−3b−3b−6bx+6bx=0⇒8ax−6bx2−6b=0⇒6bx2−8ax+6b=0⇒2(3bx2−4ax+3b)=0⇒3bx2−4ax+3b=0.\Rightarrow \dfrac{(x + 1)^2}{(x - 1)^2} = \dfrac{2a + 3b}{2a - 3b} \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{2a + 3b}{2a - 3b} \\[1em] \Rightarrow (x^2 + 1 + 2x)(2a - 3b) = (x^2 + 1 - 2x)(2a + 3b) \\[1em] \Rightarrow 2ax^2 + 2a + 4ax - 3bx^2 - 3b - 6bx = 2ax^2 + 2a - 4ax + 3bx^2 + 3b - 6bx \\[1em] \Rightarrow 2ax^2 - 2ax^2 + 2a - 2a + 4ax + 4ax - 3bx^2 - 3bx^2 - 3b - 3b - 6bx + 6bx = 0 \\[1em] \Rightarrow 8ax - 6bx^2 - 6b = 0 \\[1em] \Rightarrow 6bx^2 - 8ax + 6b = 0 \\[1em] \Rightarrow 2(3bx^2 - 4ax + 3b) = 0 \\[1em] \Rightarrow 3bx^2 - 4ax + 3b = 0.

Hence, proved that 3bx2 - 4ax + 3b = 0.

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