Question

Rationalise the denominator of 14/√108-√96+√192-√54​

Answer 1

Given :

$\frac{14}{ \sqrt{108} - \sqrt{96} + \sqrt{192} - \sqrt{54} }$

To find :

Rationalise the denominator.

Solution :

Firstly we will prime factorise all the elements of denominator, so

$\frac{14}{ \sqrt{108} - \sqrt{96} + \sqrt{192} - \sqrt{54} }$

this will become,

$\frac{14}{ \sqrt{ {2}^{2} \times {3}^{3} } - \sqrt{ {2}^{5} \times 3} + \sqrt{ {2}^{6} \times 3 } - \sqrt{2 \times {3}^{3} } }$

When we take even powers of all the elements outside the square roots,

we will get,

$\frac{14}{ 6\sqrt{ 3 } - 4\sqrt{ 6} + 8\sqrt{ 3 } - 3\sqrt{6} }$

On further solving the denominator,

we get

$\frac{14}{ 14\sqrt{ 3 } - 7\sqrt{ 6} }$

On dividing numerator and denominator by 7,

we get

$\frac{2}{ 2\sqrt{ 3 } - \sqrt{ 6} }$

then, to rationalise the denominator, we multiply numerator and denominator with :

$2 \sqrt{3 } + \sqrt{6}$

so,we will get,

$\\ \frac{2}{ 2\sqrt{ 3 } - \sqrt{ 6} } \times \frac{2 \sqrt{3} + \sqrt{6} }{2 \sqrt{3} + \sqrt{6} }$

$= \frac{4{\sqrt {3}} + 2{\sqrt {6}}} {12 - 6}\\ = \frac{4{\sqrt {3}} + 2{\sqrt {6}}} {6} \\ = \bf \frac{2 \sqrt{3} + \sqrt{6} }{3}$

So,

Rationalisation of given expression is,

$\\ \bf \: = \frac{2 \sqrt{3} + \sqrt{6} }{3}$

Answer 2

Step-by-step explanation:

answer to this question is 4