Question

5 + 2 root 3 upon 7 + 4 root 3

Answer 1

Heya friend,
Here is the answer you were looking for:
$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} }$

On rationalizing the denominator we get,

$= \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} }$

Using the identity:

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

$= \frac{5(7 - 4 \sqrt{3}) + 2 \sqrt{3}(7 - 4 \sqrt{3} ) }{ {(7)}^{2} - {(4 \sqrt{3}) }^{2} } \\ \\ = \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24 }{49 - 48} \\ \\ = 11 - 6 \sqrt{3}$

Hope this helps!!!

@Mahak24

Thanks...
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Answer 2

Heya, dear friend

Solution ⬇⬇⬇⬇⬇⤵
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$we \: \: have \: \: \\ \\ \frac{5 + 2\sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ = > \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ = > \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 8( \sqrt{3} \times \sqrt{3} ) }{( {7})^{2} - (4 \sqrt{3} {)}^{2} } \\ \\ = > \frac{(35 - 24) - 6 \sqrt{3} }{49 - 48} \\ \\ = > 11 - 6 \sqrt{3} \: \: \: \: \: \: \: answer$
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Hope it's helps you .
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