Math, asked by THEmultipleTHANKER, 5 hours ago

find p and q if \implies \dfrac{7 + \sqrt{5} }{7 - \sqrt{5} } - \dfrac{7 - \sqrt{5} }{7 + \sqrt{5} } = p - 7\sqrt{5} q

Answers

Answered by Anonymous
6

\circledcirc\mathtt\blue{ \: Question:-}

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\blue\Rsh \tt{ \frac{7 +  \sqrt{5} }{7 -  \sqrt{5}  }  -  \frac{7 -  \sqrt{5} }{7 +  \sqrt{5} }  = } \small \tt \blue{ \: p - 7 \sqrt{5}q }

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 \circledcirc\mathtt\blue{ \: Solution:-}

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\blue\Rsh \tt{ \frac{7 +  \sqrt{5} }{7 -  \sqrt{5}  }  -  \frac{7 -  \sqrt{5} }{7 +  \sqrt{5} }  = } \small \tt \blue{ \: p - 7 \sqrt{5}q }

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\Rsh\small\mathtt\blue{Rationalise \:  the \:  denominator}

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\blue\Rsh \tt{ \frac{7 +  \sqrt{5} }{7 -  \sqrt{5}  }  \times  \frac{7 +  \sqrt{5} }{7  + \sqrt{5} }  -  \frac{7 -  \sqrt{5} }{7 +  \sqrt{5} }  \times  \frac{7 -  \sqrt{5} }{7 -  \sqrt{5} }  = } \small \tt \blue{ \: p - 7 \sqrt{5} }

\blue\Rsh \tt{ \frac{(7 +  \sqrt{5}) ^{2}  }{ {(7)}^{2}  -  (\sqrt{5})^{2}   }  - \frac{(7 -  \sqrt{5})^{2}  }{ {(7)}^{2}  -  (\sqrt{5})^{2}  }  = } \small \tt \blue{ \: p - 7 \sqrt{5}q }

\blue\Rsh \tt{ \frac{ (7)^{2} +2  \times 7 \times  \sqrt{5}   +  5  }{ 49- 5   }  - \frac{((7)^{2}  - 2   \times 7 \times \sqrt{5}  +  (\sqrt{5})^{2}  )}{ 49  - 5  }  = } \small \tt \blue{ \: p - 7 \sqrt{5}q }

\blue\Rsh \tt{ \frac{ 49  + 14 \sqrt{5}  +  5  }{ 44   }  - \frac{(49 - 14 \sqrt{5}  + 5 ) }{ 44  }  = } \small \tt \blue{ \: p - 7 \sqrt{5}q }

\blue\Rsh \tt{ \frac{ 54 + 14 \sqrt{5}    }{ 44   }  - \frac{(54 - 14 \sqrt{5} )  }{ 44  }  = } \small \tt \blue{ \: p - 7 \sqrt{5} }

\blue\Rsh \tt{ \frac{ 54 + 14 \sqrt{5}    }{ 44   }  - \frac{54  +  14 \sqrt{5}  }{ 44  }  = } \small \tt \blue{ \: p - 7 \sqrt{5}q }

\blue\Rsh \tt{ \frac{ 54  - 54 + 14 \sqrt{5}    +  14 \sqrt{5}   }{ 44   }    = } \small \tt \blue{ \: p - 7 \sqrt{5}q }

\blue\Rsh \tt{ \frac{  \cancel{54  - 54} + 14 \sqrt{5}    +  14 \sqrt{5}   }{ 44   }    = } \small \tt \blue{ \: p - 7 \sqrt{5}q }

\blue\Rsh \tt { \frac{28 \sqrt{5} }{ 44   }    = } \small \tt \blue{ \: p - 7 \sqrt{5}q }

\blue\Rsh \tt { \frac{28 (\sqrt{5}) }{ 44   }    = } \small \tt \blue{ \: p - 7 \sqrt{5}q }

\blue\Rsh \tt { \frac{ \cancel{28} (\sqrt{5}) }{ \cancel{ 44}   }    = } \small \tt \blue{ \: p - 7 \sqrt{5}q }

\blue\Rsh \tt { \frac{7 (\sqrt{5}) }{ 11} =  } \small \tt \blue{ \: p - 7 \sqrt{5}q }

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\circledcirc \small\tt \blue{ \: Comparing \:  values}

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\small \boxed{\tt {\underline{p = 0}}}

\small \boxed{\tt {\underline{q =  \frac{1 }{11} }}}

Answered by LaRouge
1

Answer:

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