Question

rationalise the dinominator of (4root7-6root3)/(4root3+2root7).

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Answer 1

Answer:

$\frac{4 \sqrt{7} - 6 \sqrt{3} }{4 \sqrt{3} + 2 \sqrt{7} } \\ \\ = \frac{4 \sqrt{7} - 6 \sqrt{3} }{4 \sqrt{3} + 2 \sqrt{7} } \times \frac{4 \sqrt{3} - 2 \sqrt{7} }{4 \sqrt{3} - 2 \sqrt{7} } \\ \\ = \frac{(4 \sqrt{7} - 6 \sqrt{3}) \times (4 \sqrt{3} - 2 \sqrt{7} ) }{ {(4 \sqrt{3} )}^{2} - {( 2\sqrt{7} )}^{2} } \\ \\ = \frac{4 \sqrt{7} \times (4 \sqrt{3} - 2 \sqrt{7} ) - 6 \sqrt{3} \times (4 \sqrt{3} - 2 \sqrt{7} )}{48 - 28} \\ \\ = \frac{16 \sqrt{21} - 56 - 72 + 12 \sqrt{21} }{20} \\ \\ = \frac{28 \sqrt{21} - 122}{20}$