Math, asked by chinkudebata, 10 months ago

plz solve anyone.I can't solve plz anyone​

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amitnrw: 2b

Answers

Answered by amitnrw
5

2b   \sqrt{a^2 + 2b\sqrt{a^2 - b^2} } + \sqrt{a^2 - 2b\sqrt{a^2 - b^2} } = 2b

Step-by-step explanation:

Let say

N = \sqrt{a^2  + 2b\sqrt{a^2 - b^2} } + \sqrt{a^2  - 2b\sqrt{a^2 - b^2} }

Squaring both sides

N^2 = (\sqrt{a^2  + 2b\sqrt{a^2 - b^2} })^2  + (\sqrt{a^2  - 2b\sqrt{a^2 - b^2} })^2 + 2 (\sqrt{a^2  + 2b\sqrt{a^2 - b^2} }) (\sqrt{a^2  - 2b\sqrt{a^2 - b^2} })

N^2 =a^2  + 2b\sqrt{a^2 - b^2 }  + {a^2  - 2b\sqrt{a^2 - b^2} + 2 \sqrt{(a^2)^2  - (2b)^2(a^2 - b^2)}

N^2 =2a^2  + 2 \sqrt{(a^2)^2  - (2b)^2a^2 + (2b)^2(b^2)}\\N^2 =2a^2 + 2 \sqrt{(a^2)^2  - 2(2b^2)a^2 + (2b^2)^2} \\N^2 =2a^2 + 2 \sqrt{(2b^2 - a^2)^2} \\N^2 =2a^2 + 2(2b^2 - a^2) \\ N^2 =2a^2 + 4b^2 - 2a^2 \\N^2 = 4b^2

Taking square root both side

N = 2b

Hence

\sqrt{a^2  + 2b\sqrt{a^2 - b^2} } + \sqrt{a^2  - 2b\sqrt{a^2 - b^2} } = 2b

Learn more:

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Answered by Anonymous
8

Solution

\sqrt{ {a}^{2} + 2b \sqrt{ {a}^{2}   -  {b}^{2}}  }  + \sqrt{ {a}^{2}  -  2b \sqrt{ {a}^{2}  -  {b}^{2} } }

 \text{let }  x = \sqrt{ {a}^{2} + 2b \sqrt{ {a}^{2}  -  {b}^{2}}  }  + \sqrt{ {a}^{2}  -  2b \sqrt{ {a}^{2}   -  {b}^{2}}  }

Squaring on both sides

 \implies  {x}^{2} = \bigg( \sqrt{ {a}^{2} + 2b \sqrt{ {a}^{2}  -  {b}^{2} } }  + \sqrt{ {a}^{2}  -  2b \sqrt{ {a}^{2}  -  {b}^{2}}  } \bigg)^{2}

\implies {x}^{2} = \bigg(\sqrt{{a}^{2} + 2b \sqrt{{a}^{2} - {b}^{2}}} \bigg)^{2}+\bigg(\sqrt{ {a}^{2} - 2b \sqrt{{a}^{2} - {b}^{2}}} \bigg)^{2}\\ \\ + 2 \bigg(\sqrt{{a}^{2} + 2b\sqrt{{a}^{2} - {b}^{2}}}\bigg) \bigg(\sqrt{ {a}^{2} - 2b \sqrt{ {a}^{2} - {b}^{2} } } \bigg)

[ Because (p + q)² = p² + q² + 2pq ]

 \implies  {x}^{2} = {a}^{2} + 2b \sqrt{ {a}^{2}  -  {b}^{2}  } +  {a}^{2}  -  2b \sqrt{ {a}^{2}   -  {b}^{2}  }+ 2 \bigg( \sqrt{ ({a}^{2} + 2b \sqrt{ {a}^{2}   -  {b}^{2}})({a}^{2}  -  2b \sqrt{  {a}^{2}  -  {b}^{2}})} \bigg)

\implies  {x}^{2} = 2a^2  + 2 \bigg( \sqrt{ ({a}^{2} + 2b \sqrt{ {a}^{2}   -  {b}^{2}})({a}^{2}  -  2b \sqrt{  {a}^{2}  -  {b}^{2}})} \bigg)

\implies  {x}^{2} = 2a^2  + 2 \bigg(  \sqrt{ {(a^{2} )}^{2}  - (2b \sqrt{ {a}^{2} -  {b}^{2}  } )^{2}  } \bigg)

[ Because (p + q)(p - q) = p² - q² ]

\implies  {x}^{2} = 2a^2  + 2 \bigg(  \sqrt{ {(a^{2} )}^{2}  -  \{(2b) ^{2}  (\sqrt{ {a}^{2} -  {b}^{2}  } )^{2} \}  }  \bigg)

\implies  {x}^{2} = 2a^2  + 2 \bigg(  \sqrt{ {(a^{2} )}^{2}  -  \{(4b^{2}  ({a}^{2} -  {b}^2 ) \}  }  \bigg)

\implies  {x}^{2} = 2a^2  + 2 \bigg(  \sqrt{ {(a^{2} )}^{2}  -  \{4 {a}^{2} b^{2}  -  4{b}^4 ) \}  }  \bigg)

\implies  {x}^{2} = 2a^2  + 2 \bigg(  \sqrt{ {(a^{2} )}^{2}  -  4 {a}^{2} b^{2}   +   4{b}^4 ) \}  }  \bigg)

\implies  {x}^{2} = 2a^2  + 2 \bigg(  \sqrt{ {(a^{2} )}^{2}  -  2({a}^{2} )(2b^{2})   +   (2{b}^2)^{2} }  \bigg)

\implies  {x}^{2} = 2a^2  + 2 \bigg(  \sqrt{ {(2b^{2} )}^{2}  -  2(2{b}^{2} )(a^{2})   +   (a^2)^{2} }  \bigg)

\implies  {x}^{2} = 2a^2  + 2  \{\sqrt{ {(2b^{2} -  {a}^{2} ) }^{2} }  \}

[ Because p² - 2pq + q² = (p - q)² ]

\implies  {x}^{2} = 2a^2  + 2(2b^{2} -  {a}^{2} )

\implies  {x}^{2} = 2a^2  + 4b^{2} -  2{a}^{2}

\implies  {x}^{2} = 4b^{2}

\implies x =  \sqrt{ 4b^{2}}

\implies  \boxed{x = 2b}

Hence the answer is (C) 2b.

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