Question

Rationalize the denominator of the following
$\frac{1.6}{ \sqrt{41} - 5 }$

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{1.6}{ \sqrt{41} - 5 } \\ \\ = \frac{16}{ 10(\sqrt{41} - 5 ) } \\ \\ = \frac{16}{10 \sqrt{41} - 50 }$

On rationalizing the denominator we get,

$\frac{16}{10 \sqrt{41} - 50 } \times \frac{10 \sqrt{41} + 50 }{10 \sqrt{41} + 50 } \\ \\ = \frac{16(10 \sqrt{41} + 50) }{ {(10 \sqrt{41} )}^{2} - {(50)}^{2} } \\ \\ = \frac{160 \sqrt{41} + 800}{4100 - 2500} \\ \\ = \frac{160 \sqrt{41} + 800}{1600} \\ \\ = \frac{ \sqrt{41} + 5 }{10}$

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...
☺☺

Answer 2

hiii!!!

here's Ur answer...

$we \: have \: to \: rationalize \: \frac{1.6}{ \sqrt{41} - 5} \\ \\ = \frac{1.6}{ \sqrt{41} - 5} \times \frac{ \sqrt{41} + 5 }{ \sqrt{41} + 5} \\ \\ = \frac{1.6( \sqrt{41} + 5)}{( \sqrt{41} - 5)( \sqrt{41} + 5) } \\ \\ = \frac{1.6( \sqrt{41} + 5) }{ {( \sqrt{41}) }^{2} - {5}^{2} } \\ \\ = \frac{1.6( \sqrt{41} + 5)}{41 - 25} \\ \\ = \frac{1.6( \sqrt{41} + 5)}{16} \\ \\ = \frac{ \sqrt{41} + 5 }{10} \\ \\ hope \: this \: helps...$