Question

If root3 + root 2 / root3 - root2 = a + b root6, then find the values of a and b.

Answer 1

Heya friend,
Here is the answer you were looking for:
$\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{6}$

On rationalizing the denominator we get,

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

Using the identity :

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

$= \frac{ {( \sqrt{3} })^{2} + {( \sqrt{2} )}^{2} + 2( \sqrt{3} )( \sqrt{2} )}{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ 5 + 2 \sqrt{6} = a + b \sqrt{6} \\ \\ a = 5 \: : \: b = 2$

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

Given :

root3 + root 2 / root3 - root2 = a + b root6

i.e;  $\frac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} } = a + b\sqrt{6}$

To find :

The values of a and b

Solution :

Start with rationalizing the denominator

= $\frac{\sqrt{3} +\sqrt{2} }{\sqrt{3} -\sqrt{2} } *\frac{\sqrt{3} +\sqrt{2} }{\sqrt{3} +\sqrt{2} }$

Here, we in the numerator, apply the formula of (a+b) whole square

and in the denominator follow using a square minus b square

The formula to it is as below :

(a + b) whole sq. = a sq. + b sq. + 2ab

a sq. - b sq. = (a + b) (a - b)

On following the above rules

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

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

= $5+2\sqrt{6}$

On comparing it with $a + b\sqrt{6}$

Hence, we get ;

a = 5

b = 2

#SPJ2