Question

if 3+ root6/3-root6=a+b root6.then find the value of a and b
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Answer 1

Answer:

Step-by-step explanation:

Heya friend,

Here is the answer you were looking for:

On rationalizing the denominator we get,

Using the identity :

Hope this helps!!

Answer 2

Answer:

$\textbf{Given}:\:\boxed{ \dfrac{ 3 + \sqrt{6} }{ 3 - \sqrt{6}} = a + b\sqrt{6}}$

We have to find the values of 'a' and 'b'

Let's first start by rationalizing the LHS.

$\rightarrow \dfrac{3 + \sqrt{6}}{3 - \sqrt{6}} \times \dfrac{ 3 + \sqrt{6}}{3 + \sqrt{6}}\\\\\\\rightarrow \dfrac{ (3 + \sqrt{6} )^2}{ (3)^2 - (\sqrt{6})^2}\\\\\\\rightarrow \dfrac{ (3)^2 + 2(3)(\sqrt6) + (\sqrt{6})^2}{9 - 6}\\\\\\\rightarrow \dfrac{9 + 6 + 6\sqrt{6}}{3}\\\\\\\rightarrow \dfrac{15 + 6\sqrt{6}}{3}\\\\\\\text{Dividing by 3 we get,}\\\\\\\rightarrow 5 + 2\sqrt{6}$

Comparing the final equation with RHS we get:

$\boxed{ 5 + 2\sqrt{6}\: = a + b\sqrt{6}}\\\\\\\rightarrow \boxed{ a = 5,\:\: b = 2}$

This is the required answer.