Solve: $ \frac{\sqrt{x+a}+\sqrt{x-b}}{\sqrt{x+a}-\sqrt{x-b}}=\frac{a+b}{a-b}(a \
eq b) $

Given:

\( \frac{\sqrt{x+a}+\sqrt{x-b}}{\sqrt{x+a}-\sqrt{x-b}}=\frac{a+b}{a-b}(a \
eq b) \)

To do:

We have to solve \( \frac{\sqrt{x+a}+\sqrt{x-b}}{\sqrt{x+a}-\sqrt{x-b}}=\frac{a+b}{a-b} \).

Solution:

\( \frac{\sqrt{x+a}+\sqrt{x-b}}{\sqrt{x+a}-\sqrt{x-b}}=\frac{a+b}{a-b} \)

$ \begin{array}{l}
\frac{\left(\sqrt{x+a} +\sqrt{x-b}\right)\left(\sqrt{x+a} +\sqrt{x-b}\right)}{\left(\sqrt{x+a} -\sqrt{x-b}\right)\left(\sqrt{x+a} +\sqrt{x-b}\right)} =\frac{a+b}{a-b}\\
\frac{\left(\sqrt{x+a} +\sqrt{x-b}\right)^{2}}{\left(\sqrt{x+a}\right)^{2} -\left(\sqrt{x-b}\right)^{2}} =\frac{a+b}{a-b}\\
\frac{x+a+x-b+2\sqrt{x+a} \times \sqrt{x-b}}{x+a-( x-b)} =\frac{a+b}{a-b}\\
\frac{2x+a-b+2\sqrt{( x+a)( x-b)}}{a+b} =\frac{a+b}{a-b}\\
( a-b)\left( 2x+a-b+2\sqrt{( x+a)( x-b)}\right) =( a+b)^{2}\\
2( a-b) x+( a-b)^{2} +2( a-b)\sqrt{( x+a)( x-b)} =a^{2} +b^{2} +2ab\\
2( a-b) x+a^{2} +b^{2} -2ab+2( a-b)\sqrt{( x+a)( x-b)} =a^{2} +b^{2} +2ab\\
2( a-b) x+2( a-b)\sqrt{( x+a)( x-b)} =4ab\\
( a-b) x+( a-b)\sqrt{( x+a)( x-b)} =2ab\\
( a-b)\sqrt{( x+a)( x-b)} =2ab-( a-b) x\\
( a-b)^{2}( x+a)( x-b) =4a^{2} b^{2} +( a-b)^{2} x^{2} -4ab( a-b) x\\
( a-b)^{2}\left( x^{2} +ax-bx-ab\right) =4a^{2} b^{2} +( a-b)^{2} x^{2} -4ab( a-b) x\\
( a-b)^{3} x-ab( a-b)^{2} =4a^{2} b^{2} -4ab( a-b) x\\
x\left[( a-b)^{3} +4ab( a-b)\right] =4a^{2} b^{2} +ab( a-b)^{2}\\
x=\frac{ab\left[ 4a b +( a-b)^{2}\right]}{( a-b)\left[( a-b)^{2} +4ab\right]}\\
x=\frac{ab}{a-b}
\end{array}$ 

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Updated on: 10-Oct-2022

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