2×root5 – 3×root3 ÷ 2×root5 + 3×root3 ; rationalisethe denominater​

Question

2×root5 – 3×root3 ÷ 2×root5 + 3×root3 ; rationalisethe denominater​

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Natalia 4 weeks 2021-08-23T00:44:38+00:00 1 Answer 0 views 0

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    2021-08-23T00:45:52+00:00

    Question:-

     \to \:  \bf \:  \frac{2 \sqrt{5} - 3 \sqrt{3}  }{2 \sqrt{5}  + 3 \sqrt{3} }  \:  \:  \: rationalise \:  \:

    Solution:-

    \to \:  \bf \:  \frac{2 \sqrt{5} - 3 \sqrt{3}  }{2 \sqrt{5}  + 3 \sqrt{3} }

    \to \:  \bf \:  \frac{2 \sqrt{5} - 3 \sqrt{3}  }{2 \sqrt{5}  + 3 \sqrt{3} }  \times  \frac{2 \sqrt{5} - 3 \sqrt{3}  }{2 \sqrt{5}  - 3 \sqrt{3} }

    Using this identity

    => ( a – b )² = a² + b² – 2ab

    => (a – b )( a + b ) = ( a² – b² )

     \to \:  \bf \:  \frac{(2 \sqrt{5} - 3 \sqrt{3}  ) {}^{2} }{(2 \sqrt{5}) {}^{2} - (3 \sqrt{3}  ) {}^{2}  }

     \to \:  \bf \:  \frac{(2 \sqrt{5}) {}^{2}   + (3 \sqrt{3}) {}^{2}  - 2 \times 2 \sqrt{5} \times 3 \sqrt{3}   }{4 \times 5 - 9 \times 3}

     \bf \:  \to \:  \frac{20 + 27 - 12 \sqrt{15} }{20 - 27}

     \to \bf \:  \frac{47 - 12 \sqrt{15} }{ - 7}

     \to \bf \:  \frac{ - (47 - 12 \sqrt{15}) }{  7}

     \to \:  \bf \:  \frac{12 \sqrt{15}  - 47}{7}

     \boxed{ \green{ \bf{answer =  \:  \bf \:  \frac{12 \sqrt{15}  - 47}{7} }}}

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