Question

If √{(x-a)/(x-b)} + a/x = √{(x-b)/(x-a)} + b/x, b ≠ a , then find the value of x.

a)
$\frac{b}{a + b}$
b)
$\frac{ab}{a + b}$
c) 1

d)
$\frac{a}{a + b}$

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Answer 1

ANSWER:

Given:

• √{(x-a)/(x-b)} + a/x = √{(x-b)/(x-a)} + b/x

To Find:

• Value of x

Solution:

$\text{We are given that,}\\\\:\longrightarrow\sqrt{\dfrac{x-a}{x-b}}+\dfrac{a}{x}=\sqrt{\dfrac{x-b}{x-a}}+\dfrac{b}{x}\\\\\text{So,}\\\\:\implies\sqrt{\dfrac{x-a}{x-b}}+\dfrac{a}{x}=\sqrt{\dfrac{x-b}{x-a}}+\dfrac{b}{x}\\\\\text{On rearranging the terms,}\\\\:\implies\sqrt{\dfrac{x-a}{x-b}}-\sqrt{\dfrac{x-b}{x-a}}=\dfrac{b}{x}-\dfrac{a}{x}$

$:\implies\dfrac{\sqrt{x-a}}{\sqrt{x-b}}-\dfrac{\sqrt{x-b}}{\sqrt{x-a}}=\dfrac{b-a}{x}\\\\\text{Taking LCM in LHS,}\\\\:\implies\dfrac{(\sqrt{x-a})^2-(\sqrt{x-b})^2}{(\sqrt{x-a})(\sqrt{x-b})}=\dfrac{b-a}{x}\\\\:\implies\dfrac{x\!\!\!/-a-x\!\!\!/+b}{\sqrt{x^2-ax-bx+ab}}=\dfrac{b-a}{x}\\\\:\implies\dfrac{b-a}{\sqrt{x^2-(a+b)x+ab}}=\dfrac{b-a}{x}\\\\\text{On cancelling (b - a) on both sides,}$

$:\implies\dfrac{1}{\sqrt{x^2-(a+b)x+ab}}=\dfrac{1}{x}\\\\\text{On cross-multiplying,}\\\\:\implies x=\sqrt{x^2-(a+b)x+ab}\\\\\text{Squaring both sides,}\\\\:\implies x^2=(\sqrt{x^2-(a+b)x+ab})^2\\\\:\implies x^2\!\!\!\!\!/\:=x^2\!\!\!\!\!/\:-(a+b)x+ab\\\\:\implies 0=-(a+b)x+ab\\\\:\implies (a+b)x=ab\\\\\bf{:\implies x=\dfrac{ab}{a+b}}$

$\text{\bf{Hence, option b) \bf{\dfrac{ab}{a+b}} is the value of x.}}$