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Mathematics

If x = a+3b+a−3ba+3b−a−3b\dfrac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}}, prove that :

3bx2 - 2ax + 3b = 0

Ratio Proportion

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Answer

Given,

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

Applying componendo and dividendo,

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

Squaring both sides:

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

Hence, proved that 3bx2 - 2ax + 3b = 0.

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