Question

Rationalise the denominator of 4root 3+5root2 upon root 48 +root 18

Answer 1

Hello Mate!

$\frac{4 \sqrt{3} + 5 \sqrt{2} }{ \sqrt{48} + \sqrt{18} } \\ on \: rationalizing \\ \frac{4 \sqrt{3} + 5 \sqrt{2} }{ \sqrt{48} + \sqrt{18} } \times \frac{ \sqrt{48} - \sqrt{18} }{ \sqrt{48} - \sqrt{18} } \\ \sqrt{48} = \sqrt{2 \times 2 \times 2 \times 2 \times 3} \\ = 2 \times 2 \sqrt{3} = 4 \sqrt{3} \\ \sqrt{18} = \sqrt{2 \times 3 \times 3} \\ = 3 \sqrt{2} \\ \frac{4 \sqrt{3} + 5 \sqrt{2} }{4 \sqrt{3} + 3 \sqrt{2} } \times \frac{4 \sqrt{3} - 3 \sqrt{2} }{4 \sqrt{3} - 3 \sqrt{2} } \\ \\ \frac{(4 \sqrt{3} + 5 \sqrt{2} )(4 \sqrt{3} - 3 \sqrt{2} )}{ {(4 \sqrt{3} )}^{2} - {(3 \sqrt{2}) }^{2} } \\ \frac{16 \times 3 - 12 \sqrt{6} + 20 \sqrt{6} - 15 \times 2}{16 \times 3 - 9 \times 2} \\ \frac{48 + \sqrt{6} ( - 12 + 20) - 30}{48 - 18} \\ \frac{18 + 8 \sqrt{6} }{30} = \frac{9 + 4 \sqrt{6} }{15}$

Hope it helps☺!✌

Answer 2

Heya friend,
Here is the answer you were looking for:
$\frac{4 \sqrt{3} + 5 \sqrt{2} }{ \sqrt{48} + \sqrt{18} } \\ \\ = \frac{4 \sqrt{3} + 5 \sqrt{2} }{ \sqrt{2 \times 2 \times 2 \times 2 \times 3} + \sqrt{2 \times 3 \times 3} } \\ \\ = \frac{4 \sqrt{3} + \sqrt{2} }{ \sqrt{ {2}^{2} \times {2}^{2} \times 3 } + \sqrt{ {3}^{2} \times 2 } } \\ \\ = \frac{4 \sqrt{3} + 5 \sqrt{2} }{2 \times 2 \sqrt{3} + 3 \sqrt{2} } \\ \\ = \frac{4 \sqrt{3} + 5 \sqrt{2} }{4 \sqrt{3} + 3 \sqrt{2} }$

On rationalizing the denominator we get,

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

Using the identity :

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


$= \frac{4 \sqrt{3}(4 \sqrt{3} - 3 \sqrt{2} ) + 5 \sqrt{2}(4 \sqrt{3} - 3 \sqrt{2} ) }{ {(4 \sqrt{3} )}^{2} - {(3 \sqrt{2}) }^{2} } \\ \\ = \frac{16 \times 3 - 12 \sqrt{6} + 20 \sqrt{6} - 15 \times 2 }{48 - 18} \\ \\ = \frac{48 - 30 - 12 \sqrt{6} + 20 \sqrt{6} }{30} \\ \\ = \frac{18 + 8 \sqrt{6} }{30} \\ \\ = \frac{9 + 4\sqrt{6} }{15}$

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...
☺☺