Question

If 'a' and 'b' are rational numbers, find the value of a and b of V3+v2/v3-v2= a+bv6​

Answer 1

Step-by-step explanation:

$\bf \underline{Solution-}$

Give expression

$\rm \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = a + b \sqrt{3}$

The denominator is √3 - √2.

We know that

Rationalising factor of √a - √b = √a + √b.

So, the rationalising factor of √3-√2 = √3+√2.

On rationalising the denominator them

$\rm \longrightarrow \: \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$\rm \longrightarrow \: \frac{( \sqrt{3} + \sqrt{2} )( \sqrt{3} + \sqrt{2}) }{ ( \sqrt{3} - \sqrt{2})( \sqrt{3} + \sqrt{2} )}$

Now, applying algebraic identity in denominator because it is in the form of:

(a - b) (a + b) = a² - b²

Where, we have to put in our expression, a = √3 and b = √2.

$\rm \longrightarrow \: \frac{( \sqrt{3} + \sqrt{2})( \sqrt{3} + \sqrt{2} ) }{( \sqrt{3} {)}^{2} - ( \sqrt{2} {)}^{2} }$

$\rm \longrightarrow \: \frac{( \sqrt{3} + \sqrt{2} )( \sqrt{3} + \sqrt{2} )}{3 - 2}$

$\rm \longrightarrow \: \frac{( \sqrt{3} + \sqrt{2} )( \sqrt{3} + \sqrt{2} )}{1}$

$\longrightarrow \: ( \sqrt{3} + \sqrt{2} )( \sqrt{3} + \sqrt{2} )$

Now, applying algebraic identity because this term in the form of:

(a + b) (a +b) = a² + 2ab + b²

Where, we have to put in our expression, a = √3 and b = √2.

$\rm \longrightarrow \: ( \sqrt{3} + \sqrt{2} {)}^{2}$

$\rm \longrightarrow \: ( \sqrt{3} {)}^{2} + 2 \sqrt{3 \times 2} + ( \sqrt{2} {)}^{2}$

$\rm \longrightarrow \: 3 + 2 \sqrt{3 \times 2} + 2$

$\rm \longrightarrow \: 5 + 2 \sqrt{3 \times 2}$

$\rm \longrightarrow \: 5 + 2 \sqrt{6}$

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

On comparing with the value of R.HS

we have, a = 5 and b = 2√6 = 2.