Question

rationalize it!

Right answer with full solution will get brainlisted​

Answer 1

Answer:-

⟹ $\frac{3+√2}{3-√2}$

⟹$\frac{3+√2}{3-√2}$ × $\frac{3+√2}{3+√2}$

⟹ $\frac{(3+√2)²}{3²-(√2)²}$

⟹ $\frac{3²+2+6√2}{9-2}$

⟹$\frac{9+2+6√2}{7}$

⟹ $\frac{11+6√2}{7}$

Answer 2

Answer:

$⟹ \frac{11 + 6 \sqrt{2} }{7} \: \: \tt \red{Ans }.$

Step-by-step explanation:

Given that:-

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

What to do:-

To rationalise the denominator.

Solution:-

We have,

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

The denominator is 3-√2. Multiplying the numerator and denominator by 3+√2,

We get

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

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

⬤ Applying Algebraic Identity

(a+b)(a+b) = (a+b)² = a²+2ab+b² to the numerator and

(a-b)(a+b) = a² - b² to the denominator

We get,

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

$⟹ \frac{ {3}^{2} + 2 \times 3 \sqrt{2} + ( \sqrt{2} {)}^{2} }{9 - 2}$

$⟹ \frac{9 + 6 \sqrt{2} + 2}{7}$

$⟹ \frac{11 + 6 \sqrt{2} }{7} \: \: \tt \red{Ans }.$

Hence, the denominator is rationalised.

Know more Algebraic Identities:-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc

I hope it's help with...☺