# If x=2ab/b^2+1 Find the value of ...

If x=2ab/b^2+1 Find the value of (a+x)^1/2+ (a-x)^1/2/(a+x)^1/2- (a-x)^1/2

• 1 answers

Rashmi Bajpayee 1 year, 9 months ago

Given: {tex}x = {{2ab} \over {{b^2} + 1}}{/tex}

{tex}{{\sqrt {a + x} + \sqrt {a - x} } \over {\sqrt {a + x} - \sqrt {a - x} }}{/tex}

= {tex}{{\sqrt {a + x} + \sqrt {a - x} } \over {\sqrt {a + x} - \sqrt {a - x} }} \times {{\sqrt {a + x} + \sqrt {a - x} } \over {\sqrt {a + x} + \sqrt {a - x} }}{/tex}

= {tex}{{{{\left( {\sqrt {a + x} } \right)}^2} + {{\left( {\sqrt {a - x} } \right)}^2} + 2\sqrt {a + x} .\sqrt {a - x} } \over {{{\left( {\sqrt {a + x} } \right)}^2} - {{\left( {\sqrt {a - x} } \right)}^2}}}{/tex}

= {tex}{{a + x + a - x + 2\sqrt {{a^2} - {x^2}} } \over {a + x - a + x}}{/tex}

= {tex}{{2a + 2\sqrt {{a^2} - {x^2}} } \over {2x}}{/tex}

= {tex}{{a + \sqrt {{a^2} - {x^2}} } \over x}{/tex}

= {tex}{{a + \sqrt {{a^2} - {{\left( {{{2ab} \over {{b^2} + 1}}} \right)}^2}} } \over {{{2ab} \over {{b^2} + 1}}}}{/tex}

= {tex}a + \sqrt {{a^2} - {{4{a^2}{b^2}} \over {{{\left( {{b^2} + 1} \right)}^2}}}} \times {{{b^2} + 1} \over {2ab}}{/tex}

= {tex}a + a\sqrt {{{{{\left( {{b^2} + 1} \right)}^2} - 4{b^2}} \over {{{\left( {{b^2} + 1} \right)}^2}}}} \times {{{b^2} + 1} \over {2ab}}{/tex}

= {tex}a\left[ {1 + {1 \over {{b^2} + 1}}\sqrt {{b^4} + 1 + 2{b^2} - 4{b^2}} } \right] \times {{{b^2} + 1} \over {2ab}}{/tex}

= {tex}{{{b^2} + 1 + \sqrt {{b^4} + 1 - 2{b^2}} } \over {{b^2} + 1}} \times {{{b^2} + 1} \over {2b}}{/tex}

= {tex}{{{b^2} + 1 + \sqrt {{{\left( {{b^2} - 1} \right)}^2}} } \over {2b}}{/tex}

= {tex}{{{b^2} + 1 + {b^2} - 1} \over {2b}}{/tex}

= {tex}{{2{b^2}} \over {2b}}{/tex}

= {tex}b{/tex}

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