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Mathematics

If x = paba+b\dfrac{pab}{a + b}, prove that

x+pax−pa−x+pbx−pb=2(a2−b2)ab.\dfrac{x + pa}{x - pa} - \dfrac{x + pb}{x - pb} = \dfrac{2(a^2 - b^2)}{ab}.

Ratio Proportion

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Answer

Given,

x=paba+b⇒xpa=ba+b and xpb=aa+bx = \dfrac{pab}{a + b} \\[1em] \Rightarrow \dfrac{x}{pa} = \dfrac{b}{a + b} \text{ and } \dfrac{x}{pb} = \dfrac{a}{a + b}

Applying componendo and dividendo on both equations,

⇒x+pax−pa=b+a+bb−a−b and x+pbx−pb=a+a+ba−a−b⇒x+pax−pa=−2b+aa and x+pbx−pb=−2a+bb\Rightarrow \dfrac{x + pa}{x - pa} = \dfrac{b + a + b}{b - a - b} \text{ and } \dfrac{x + pb}{x - pb} = \dfrac{a + a + b}{a - a - b} \\[1em] \Rightarrow \dfrac{x + pa}{x - pa} = -\dfrac{2b + a}{a} \text{ and } \dfrac{x + pb}{x - pb} = -\dfrac{2a + b}{b}

Subtracting both the equations,

⇒x+pax−pa−x+pbx−pb=−2b+aa−(−2a+bb)=−2b+aa+2a+bb=−2b2−ab+2a2+abab=2(a2−b2)ab=R.H.S.\Rightarrow \dfrac{x + pa}{x - pa} - \dfrac{x + pb}{x - pb} = -\dfrac{2b + a}{a} - \Big(-\dfrac{2a + b}{b}\Big) \\[1em] = -\dfrac{2b + a}{a} + \dfrac{2a + b}{b} \\[1em] = \dfrac{-2b^2 - \cancel{ab} + 2a^2 + \cancel{ab}}{ab} \\[1em] = \dfrac{2(a^2 - b^2)}{ab} = \text{R.H.S.} \\[1em]

Since, L.H.S. = R.H.S. , hence, proved that,

x+pax−pa−x+pbx−pb=2(a2−b2)ab.\dfrac{x + pa}{x - pa} - \dfrac{x + pb}{x - pb} = \dfrac{2(a^2 - b^2)}{ab}.

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