Question

rationalise 3-√2/3+√2​

Answer 1

GIVEN :-

$\\ \sf \: \dfrac{3 - \sqrt{2} }{3 + \sqrt{2} }$

SOLUTION :-

We have ,

$\\ \sf \: \dfrac{3 - \sqrt{2} }{3 + \sqrt{2} }$

Now , we will rationalise the denominator.

Rationalising factor is (3 - √2).

We will multiply and divide (3 - √2) to the given fraction.

$\\ \implies \sf \: \dfrac{3 - \sqrt{2} }{3 + \sqrt{2} } \times \frac{3 - \sqrt{2} }{3 - \sqrt{2} } \\ \\ \implies \sf \: \dfrac{( {3 - \sqrt{2} )}^{2} }{(3 + \sqrt{2} )(3 - \sqrt{2}) }$

In numerator,

★ (a - b)² = a² + b² - 2ab

Here ,

• a = 3
• b = √2

______________

In denominator,

★ (a+b)(a-b) = a² - b²

Here ,

• a = 3
• b = √2

Putting values,

$\\ \implies \sf \dfrac{ {3}^{2} + {( \sqrt{2}) }^{2} - 2(3)( \sqrt{2} ) }{ {3}^{2} - ( { \sqrt{2} )}^{2} } \\ \\ \\ \implies \sf \: \dfrac{9 + 2 - 6 \sqrt{2} }{9 - 4} \\ \\ \\ \implies \sf \: \dfrac{11 - 6 \sqrt{2} }{5}$

Hence ,

$\\ \underline{\boxed{\sf \: \dfrac{3 - \sqrt{2} }{3 + \sqrt{2} } = \frac{11 - 6 \sqrt{2} }{5} }}$

MORE IDENTITIES :-

★ (a+x)(a+y) = a² + (x+y)a + xy

★ (a+b)² = a² + b² + 2ab

★ (a+b)³ = a³ + 3a²b + 3ab² + b³

★ (a-b)³ = a³ - 3a²b + 3ab² - b³