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Mathematics

If x=b+3a+b−3ab+3a−b−3ax = \dfrac{\sqrt{b + 3a} + \sqrt{b - 3a}}{\sqrt{b + 3a} - \sqrt{b - 3a}}, prove that : 3ax2 - 2bx + 3a = 0.

Ratio Proportion

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Answer

Given,

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

Applying componendo and dividendo,

⇒x+1x−1=b+3a+b−3a+b+3a−b−3ab+3a+b−3a−(b+3a−b−3a)⇒x+1x−1=b+3a+b−3a+b+3a−b−3ab+3a+b−3a−b+3a+b−3a⇒x+1x−1=2b+3a2b−3a⇒x+1x−1=b+3ab−3a\Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{b + 3a} + \sqrt{b - 3a} + \sqrt{b + 3a} - \sqrt{b - 3a}}{\sqrt{b + 3a} + \sqrt{b - 3a} - (\sqrt{b + 3a} - \sqrt{b - 3a})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{b + 3a} + \sqrt{b - 3a} + \sqrt{b + 3a} - \sqrt{b - 3a}}{\sqrt{b + 3a} + \sqrt{b - 3a} - \sqrt{b + 3a} + \sqrt{b - 3a}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{b + 3a}}{2\sqrt{b - 3a}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{b + 3a}}{\sqrt{b - 3a}}

Squaring both sides:

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

Hence, proved that 3ax2 - 2bx + 3a = 0.

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