Question

Rationalise denominator :-
1) 3+√2/4√2
2)2+√3/2-√3
3)16/√41-5
4)3√5+√3/√5-√3

Answer 1

Hey friend,
Here is the answer you were looking for:

$(1) \frac{3 + \sqrt{2} }{4 \sqrt{2} }$

On rationalizing the denominator we get,

$= \frac{3 + \sqrt{2} }{4 \sqrt{2} } \times \frac{4 \sqrt{2} }{4 \sqrt{2} } \\ \\ = \frac{3 \times 4 \sqrt{2} + \sqrt{2} \times 4 \sqrt{2} }{ {(4 \sqrt{2} })^{2} } \\ \\ = \frac{12 \sqrt{2} + 8 }{32} \\ \\ = \frac{4(3 \sqrt{2} + 2)}{4(8)} \\ \\ = \frac{3 \sqrt{2} + 2}{8}$

$(2) \frac{2 + \sqrt{3} }{2 - \sqrt{3} }$
On rationalizing the denominator we get,

$\frac{2 + \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} }$

Using the identity :

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

$= \frac{ {(2)}^{2} + {( \sqrt{3} )}^{2} + 2(2)( \sqrt{3}) }{ {(2)}^{2} - {( \sqrt{3}) }^{2} } \\ \\ = \frac{4 + 3 + 4 \sqrt{3} }{4 - 3} \\ \\ = 7 + 4 \sqrt{3}$

$(3) \frac{16}{ \sqrt{41} - 5}$

On rationalizing the denominator we get,

$\frac{16}{ \sqrt{41} - 5 } \times \frac{ \sqrt{41} + 5}{ \sqrt{41} + 5}$

Using the identity

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

$\frac{16 \times \sqrt{41} + 16 \times 5}{ {( \sqrt{41} )}^{2} - {(5)}^{2} } \\ \\ = \frac{16 \sqrt{41} + 80}{41 - 25} \\ \\ = \frac{16 \sqrt{41} + 80 }{16} \\ \\ = \frac{16( \sqrt{41} + 5) }{16} \\ \\ = \sqrt{41} + 5$
$(4) \frac{3 \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

On rationalizing the denominator we get,

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

Using same identity is above

$= \frac{3 \sqrt{5} ( \sqrt{5} + \sqrt{3}) + \sqrt{3} ( \sqrt{5} + \sqrt{3} )}{ {( \sqrt{5}) }^{2} - {( \sqrt{3} )}^{2} } \\ \\ = \frac{15 + 3 \sqrt{15} + \sqrt{15} + 3}{5 - 3} \\ \\ = \frac{18 + 4 \sqrt{15} }{2} \\ \\ = \frac{2(9 + 2 \sqrt{15}) }{2} \\ \\ = 9 + 2 \sqrt{15}$

Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
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