Math, asked by reetaraviagrawal, 6 hours ago

answer fast plz plea​

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Answers

Answered by Anonymous
13

Given equation:

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

To find:

  • Value of a and b

Solution:

By rationalising the LHS of given equation, (i.e. by multiplying both numerator and denominator of LHS with the rationalising factor which is 3 + √7), we get:

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

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

‎ ‎ ‎

Now, use the following identities:

  • \boxed{\tt (A+B)^2 = A^2 + B^2 + 2AB}
  •  \boxed{\tt (A+B)(A-B) = A^2 - B^2}

‎ ‎ ‎

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

\sf \implies\dfrac{9  + 7  + 6\sqrt{7} }{9 - 7} =  a + b \sqrt{7}

\sf \implies\dfrac{16  + 6\sqrt{7} }{2} =  a + b \sqrt{7}

\sf \implies\dfrac{ \not2(8  + 3\sqrt{7}) } {\not2} =  a + b \sqrt{7}

\sf \implies8  + 3\sqrt{7} =  a + b \sqrt{7}

‎ ‎ ‎

On comparing LHS and RHS, we get:

  • a = 8
  • b = 3

Therefore, option (c) is correct.

Answered by MathCracker
11

Answer :-

  • Hence, option (c) is correct.

Step by step explanation :-

Given expression :

  \tt{\frac{3 +  \sqrt{7} }{3 -  \sqrt{7}  }  = a + b \sqrt{7} } \\

First we rationalize the denominator,

\rm:\longmapsto{ \frac{3 +  \sqrt{7} }{3 -  \sqrt{7} }  \times  \frac{3 +  \sqrt{7} }{3  +  \sqrt{7} }  } \\  \\\rm:\longmapsto{ \frac{(3 +  \sqrt{7}) {}^{2}  }{(3 -  \sqrt{7})(3 +  \sqrt{7} ) } }

Using (a+b)² identity in numerator and (a+b) (a-b) identity use in denominator,

\rm:\longmapsto{ \frac{(3) {}^{2} + 2(3)( \sqrt{7} ) + ( \sqrt{7}) {}^{2}   }{(3) {}^{2} - ( \sqrt{7} ) {}^{2}  } } \\  \\ \rm:\longmapsto{ \frac{9 + 6 \sqrt{7}  + 7}{9 - 7} } \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\\rm:\longmapsto{ \frac{16 + 6 \sqrt{7} }{2} =  \frac{ \cancel2(8 + 3 \sqrt{7})}{ \cancel2}  }  \\  \\ \rm:\longmapsto{8 + 3 \sqrt{7} } = a + b \sqrt{7}  \:  \:  \:  \:  \:  \:  \:  \:  \:

Comparing LHS and RHS we get,

  • a = 8
  • b = 3

Hence, option (c) is correct.

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