Question

Rationalise the denominator

$\frac{2}{\sqrt{5} +\sqrt{3} +2}$

Answer 1

Answer:

To rationalize the denominator, you must multiply both the numerator and the denominator by the conjugate of the denominator. Remember to find the conjugate all you have to do is change the sign between the two terms. Step 2: Distribute (or FOIL) both the numerator and the denominator.

Answer 2

Step-by-step explanation:

Given :

$\frac{2}{ \sqrt{5} + \sqrt{3} + 2 }$

Solution :

$= > \frac{2}{ \sqrt{5} + \sqrt{3} + 2 } \times \frac{( \sqrt{5} + \sqrt{3} ) - 2}{( \sqrt{5} + \sqrt{3}) - 2 }$

$= > \frac{2( \sqrt{5} + \sqrt{3} - 2) }{( \sqrt{5} + \sqrt{3} ) ^{2} - {2}^{2} }$

$= > \frac{2( \sqrt{5} + \sqrt{3} - 2)}{5 + 3 + 2 \sqrt{5} \sqrt{3} - 4}$

$= > \frac{2( \sqrt{5} + \sqrt{3} - 2)}{4 + 2 \sqrt{15} }$

$= > \frac{ \sqrt{5} + \sqrt{3} - 2 }{2 + \sqrt{15} }$

$= > \frac{ \sqrt{5} + \sqrt{3} - 2 }{2 + \sqrt{15} } \times \frac{2 - \sqrt{15} }{2 - \sqrt{15} }$

$= > \frac{2 \sqrt{5} + 2 \sqrt{3} - 4 - \sqrt{5} \sqrt{15} - \sqrt{3} \sqrt{15} + 2 \sqrt{15} }{ {2}^{2} - ( \sqrt{15} ) ^{2} }$

$= > \frac{2 \sqrt{5} + 2 \sqrt{3} - 4 - 5 \sqrt{3} - 3 \sqrt{5} + 2 \sqrt{15} }{4 - 15}$

$= > \frac{ - \sqrt{5} - 3 \sqrt{3} - 4 + 2 \sqrt{15} }{ - 11}$

$= > \frac{ \sqrt{5} + 3 \sqrt{3} - 2 \sqrt{15} + 4 }{11} .$

Hence :

The rationalise denominator is 11.

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