Question

rationalise
1) root6/root2 +root 3

Answer 1

Identity used :

$(a + b)(a - b) = {a}^{2} - {b}^{2}$

Now,

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

On rationalizing the denominator we get,

$= \frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} } \\ \\ = \frac{ \sqrt{6} ( \sqrt{2} - \sqrt{3} )}{ {( \sqrt{2}) }^{2} - {( \sqrt{3} )}^{2} } \\ \\ = \frac{ \sqrt{12} + \sqrt{18} }{2 - 3} \\ \\ = \frac{ \sqrt{2 \times 2 \times 3} + \sqrt{3 \times 3 \times 2} }{ - 1} \\ \\ = - ( \sqrt{ {2}^{2} \times 3} + \sqrt{ {3}^{2} \times 2 } ) \\ \\ = - (2 \sqrt{3} + 3 \sqrt{2} ) \\ \\ = - 2 \sqrt{3} - 3 \sqrt{2}$

Answer 2

Identity used :

(a + b)(a - b) = {a}^{2} - {b}^{2}(a+b)(a−b)=a

2

−b

2

Now,

\begin{lgathered}\frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \\\end{lgathered}

2

+

3

6

On rationalizing the denominator we get,

\begin{lgathered}= \frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} } \\ \\ = \frac{ \sqrt{6} ( \sqrt{2} - \sqrt{3} )}{ {( \sqrt{2}) }^{2} - {( \sqrt{3} )}^{2} } \\ \\ = \frac{ \sqrt{12} + \sqrt{18} }{2 - 3} \\ \\ = \frac{ \sqrt{2 \times 2 \times 3} + \sqrt{3 \times 3 \times 2} }{ - 1} \\ \\ = - ( \sqrt{ {2}^{2} \times 3} + \sqrt{ {3}^{2} \times 2 } ) \\ \\ = - (2 \sqrt{3} + 3 \sqrt{2} ) \\ \\ = - 2 \sqrt{3} - 3 \sqrt{2}\end{lgathered}

=

2

+

3

6

×

2

−

3

2

−

3

=

(

2

)

2

−(

3

)

2

6

(

2

−

3

)

=

2−3

12

+

18

=

−1

2×2×3

+

3×3×2

=−(

2

2

×3

+

3

2

×2

)

=−(2

3

+3

2

)

=−2

3

−3

2