Question

rationalize the dinominator of root 3+root 2/root3- root2.

Answer 1

Hi ,

( √3 + √2 ) / ( √3 - √2 )

= [ (√3+√2)(√3+√2)]/[√3-√2)(√3+√2)]

= ( √3 + √2 )² / [ ( √3 )² - ( √2 )² ]

= [ (√3)² + ( √2)²+2×√3×√2 ] /( 3 - 2 )

[ By using ( a + b )² = a² + b² + 2ab ]

= ( 3 + 2 + 2√6 ) / 1

= 5 + 2√6

I hope this helps you.

: )

Answer 2

Hey friends, Harish here.

Here is your answer:

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

To rationalize the denominator, Multiply and divide the number by the conjugate. 

The Conjugate of  Denominator is ---- √3 + √2.

Then, 
 
$\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3}- \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3}+\sqrt{2} }$

(Here, $[(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2}) = (\sqrt{3}^{2}-\sqrt{2}^{2}) = 3-2]$)

We get that by using the identity (a-b)(a+b) = a² - b².

Then,

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

⇒$\frac{( \sqrt{3}+ \sqrt{2})^{2} }{1}$

Therefore the denominator is rationalized.
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Hope my answer is helpful to u.