4. 3+/6 √3+√2 = a + b V3 5. Find a and b if​

Question

4.
3+/6
√3+√2
= a + b V3
5. Find a and b if​

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Emery 3 weeks 2021-11-06T22:37:15+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-11-06T22:39:05+00:00

     \bf \frac{3 +  \sqrt{6} }{ \sqrt{3} +  \sqrt{2}  }  = a + b \sqrt{3}   \\

    • To figure out the value of a and b we have to rationalize the LHS,

    Rationalizing the LHS, we get,

     =  \frac{3 +  \sqrt{6} }{ \sqrt{3}  +  \sqrt{2} }  \times  \frac{ \sqrt{3} -  \sqrt{2}  }{ \sqrt{3} - \sqrt{2}  }  \\  \\  =  \frac{3 +  \sqrt{6}( \sqrt{3} -  \sqrt{2}  )  }{ ( \sqrt{3}  +   \sqrt{2}  ) ( \sqrt{3} -  \sqrt{2}  )  }  \\

    • In the denominator, an identity is applicable, (a+b)(a-b) = a² – b²

     =  \frac{3( \sqrt{3} -  \sqrt{2}  )  +  \sqrt{6} ( \sqrt{3} -  \sqrt{2}  ) }{ {( \sqrt{3} )}^{2} -  { (\sqrt{2}) }^{2}  }  \\  \\  =  \frac{3 \sqrt{3} - 3 \sqrt{2} +  \sqrt{18}   -  \sqrt{12}  }{3 - 2}  \\  \\  =  \frac{3 \sqrt{3} + 3 \sqrt{2} + 3 \sqrt{2}  - 2 \sqrt{3}   }{1}  \\  \\  = 3 \sqrt{3} + 3 \sqrt{2} + 3 \sqrt{2}  - 2 \sqrt{3} \\  =  (3 - 2)\sqrt{3}  + (3 + 3) \sqrt{2}  \\  =   \sqrt{3}  + 6 \sqrt{2} \:  \:  \: \rm \:  or \:  \: \red{ 6 \sqrt{2}  +  \sqrt{3} }

    Thus,

    • a = 6√2 & b = 1
    0
    2021-11-06T22:39:11+00:00

    Answer:

    question is not clear please resend

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