Math, asked by khushikushwaha, 4 months ago

find the value of a and b: √7+√3÷√7-√3=a-b√21​

Answers

Answered by MaIeficent
14

Step-by-step explanation:

Given:-

  • \sf \dfrac{ \sqrt{7} +  \sqrt{3}  }{ \sqrt{7} -  \sqrt{3}  }  = a - b \sqrt{21}

To Find:-

  • The value of a and b

Solution:-

\sf \implies \dfrac{ \sqrt{7} +  \sqrt{3}  }{ \sqrt{7} -  \sqrt{3}  }  = a - b \sqrt{21}

By rationalizing the denominator:-

\sf  \implies  \dfrac{ \sqrt{7} +  \sqrt{3}  }{ \sqrt{7} -  \sqrt{3}  } \times  \dfrac{ \sqrt{7}   +  \sqrt{3}  }{ \sqrt{7}  + \sqrt{3}  }   = a - b \sqrt{21}

\sf  \implies  \dfrac{ (\sqrt{7} +  \sqrt{3})( \sqrt{7}  +  \sqrt{3}) }{ (\sqrt{7} -  \sqrt{3}) ( \sqrt{7}  +  \sqrt{3}) }    = a - b \sqrt{21}

\sf  \implies  \dfrac{ (\sqrt{7} +  \sqrt{3}) ^{2}  }{ (\sqrt{7})^{2}  -  (\sqrt{3})^{2}  }    = a - b \sqrt{21}

\sf  \implies  \dfrac{ (\sqrt{7}) ^{2}  +  (\sqrt{3})^{2}  + 2( \sqrt{7})( \sqrt{3} )  }{ 7 - 3}    = a - b \sqrt{21}

\sf  \implies  \dfrac{ 7 +  3  + 2 \sqrt{21} }{7 - 3}    = a - b \sqrt{21}

\sf  \implies  \dfrac{ 10 + 2 \sqrt{21} }{4}    = a - b \sqrt{21}

\sf  \implies  \dfrac{ 2(5+  \sqrt{21}) }{ 4}    = a - b \sqrt{21}

\sf  \implies  \dfrac{ 5+  \sqrt{21} }{2}    = a - b \sqrt{21}

\sf  \implies  \dfrac{ 5}{2}+   \dfrac{\sqrt{21} }{2}    = a - b \sqrt{21}

Comparing LHS and RHS:-

\sf  \implies  a =  \dfrac{5}{2} \:  \:  \: and \:  \:  \: b =    - \dfrac{1}{2}

\sf  \dashrightarrow\underline{  \boxed{  \sf\therefore \:   a =  \dfrac{5}{2} \:  \:  \: and \:  \:  \: b =    - \dfrac{1}{2}}}

Answered by tarracharan
5

Answer :-

\boxed{\bold{\red{a =\dfrac{5}{2}}}} ; \boxed{\bold{\purple{b =-\dfrac{1}{2}}}}

Given :-

\sf{\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} = a - b\sqrt{21}}

To find :-

• The value of ‘a’ and ‘b’.

Method of solving :-

• Rationalizing LHS.

Solution :-

\sf{LHS = \dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}}

\sf{LHS = \dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} \times  \dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}}}

\sf{LHS = \dfrac{(\sqrt{7}+\sqrt{3})²}{(\sqrt{7})²-(\sqrt{3})²}}

\sf{LHS = \dfrac{(\sqrt{7})²+(\sqrt{3})²+2(\sqrt{7})(\sqrt{3})}{7-3}}

\sf{LHS = \dfrac{7+3+2(\sqrt{21})}{4}}

\sf{LHS = \dfrac{10+2(\sqrt{21})}{4}}

\sf{LHS = \dfrac{5}{2}+\dfrac{1}{2}\sqrt{21}}

\sf{\dfrac{5}{2}+\dfrac{1}{2}\sqrt{21} = a-b\sqrt{21}}

\bold{\red{a =\dfrac{5}{2}}} ; \bold{\purple{b =-\dfrac{1}{2}}}

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