Question

Find the value of a and b:
$\frac{3 + \sqrt{6} }{ \sqrt{3 } + \sqrt{2} } = a + b \sqrt{3}$
​

Answer 1

Answer:

a = 0

b = 3

Step-by-step explanation:

As per the provided information in the given question, we have :

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

We are asked to calculate the value of a and b.

In order to calculate the value of a and b, firstly we need to rationalize the denominator. In order to rationalise the denominator, we multiply the numerator and the denominator with the rationalising factor of the denominator.

Here, the denominator is in the form of (a + b). Rationalising factor of (a + b) is (a - b). So, rationalising factor of (√3 + √2) is (√3 - √2). We'll multiply (√3 - √2) with both the numerator and the denominator to rationalise the denominator.

Solving L.H.S :

$\longmapsto \rm { \dfrac{3 + \sqrt{6} }{ \sqrt{3 } + \sqrt{2} } \times \dfrac{\sqrt{3} - \sqrt{2} }{\sqrt{3} - \sqrt{2}} }$

Multiplying (√3 - √2) with both the numerator and the denominator.

$\longmapsto \rm { \dfrac{(\sqrt{3} - \sqrt{2} )(3 + \sqrt{6} ) }{ (\sqrt{3} - \sqrt{2} )(\sqrt{3 } + \sqrt{2} )} }$

Rearranging the terms.

$\longmapsto \rm { \dfrac{\sqrt{3}(3 + \sqrt{6}) - \sqrt{2}(3 + \sqrt{6} ) }{(\sqrt{3})^2 - (\sqrt{2})^2} }$

Used the identity,

• $\bf (a +b)(a - b) = a^2 - b^2$

$\longmapsto \rm { \dfrac{3 \sqrt{3} + \sqrt{18} - 3\sqrt{2} - \sqrt{12} }{3 - 2} }$

Performing multiplication and writing the squares of the numbers in the denominator.

$\longmapsto \rm { \dfrac{3 \sqrt{3} + 3\sqrt{2}- 3\sqrt{2} - 2\sqrt{3}}{1 }}$

Performing subtraction in the numerator.

$\longmapsto \rm { 3 \sqrt{3} + 3\sqrt{2}- 3\sqrt{2} - 2\sqrt{3} }$

$\longmapsto \rm { 3 \sqrt{3} - 2\sqrt{3} }$

Taking √3 as common.

$\longmapsto \rm {( 3-2) \sqrt{3} }$

Performing subtraction.

$\longmapsto \rm {1 \sqrt{3} }$

We can write it as,

$\longmapsto \bf {\sqrt{3} }$

Now, comparing L.H.S and R.H.S, we get :

$\longmapsto \bf{\red{0} + \red{1} \sqrt{3} = \red{a} +\red{ b}\sqrt{3}}$

∴ Value of a is 0 and value of b is 3.