Question

root3 + root2 by root3 - root2 = a+b root6

Answer 1

Answer:

$\large{\underline{\boxed{\sf a = 5 \: \:and \: \: b = 2}}}$

Step-by-step explanation:

Given that ;

$\sf \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{6}$

Consider LHS,

$\sf \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ \sf \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2}}{ \sqrt{3} + \sqrt{2} } \\ \\ \sf \frac{( \sqrt{3} + \sqrt{2} ) {}^{2} }{( \sqrt{3})^{2} - (\sqrt{2}) {}^{2} } \\ \\ \sf \frac{( \sqrt{3}) {}^{2} + 2 \times \sqrt{3} \times \sqrt{2} + ( \sqrt{2} ) {}^{2} }{3 - 2} \\ \\ \sf \frac{3 + 2 \sqrt{6} + 2}{1} \\ \\ \sf 5 + 2 \sqrt{6}$

On comparing LHS with RHS,

$\sf 5 + 2 \sqrt{6} = a + b \sqrt{6}$

We get;

• a = 5
• b = 2

Hence, the value of a and b is 5 and 2 respectively.

Answer 2

Given:-

$\dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{6}$

Answer:-

a = 5

b = 2

Now, solving L. H. S

we get,

$= \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

Using suitable identity,

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

$= \dfrac{( \sqrt{3} + \sqrt{2} ) ^{2} }{ ({ \sqrt{3} })^{2} - (\sqrt{2})^{2} }$

$= \dfrac{( \sqrt{3})^{2} + { (\sqrt{2}) }^{2} + 2. \sqrt{3} . \sqrt{2} }{ {( \sqrt{3}) }^{2} - { (\sqrt{2}) }^{2} }$

$= \dfrac{3 + 2 + 2 \sqrt{6} }{3 - 2}$

$= \dfrac{5 + 2 \sqrt{6} }{1}$

$= 5 + 2 \sqrt{6}$

Now , compare with R. H. S

$5 + 2 \sqrt{6} = a + b \sqrt{6}$

we get,

$a = 5 \: \: \: \: b \: = 2$