Question

find a and b if 3+√2/3 -√2=a+b√2

Answer 1

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

=====================

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


By rationalization,


$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } \times \frac{3 + \sqrt{2} }{3 + \sqrt{2} } \\ \\ \\ \frac{(3 + \sqrt{2} )^{2} }{ {(3)}^{2} - ( \sqrt{2} )^{2} } \\ \\ \\ \frac{9 + 2 + 6 \sqrt{2} }{9 - 4} \\ \\ \\ \frac{11 + 6 \sqrt{2} }{5}$


Comparing values,


$\frac{11 + 6 \sqrt{2} }{5} = a + b \sqrt{2}$



$a = \frac{11}{5} \\ \\ \\ b = \frac{6}{5}$




I hope this will help you


(-:

Answer 2

Heya ✋

Let see your answer !!!!!

Given that

3 + √2 / 3 - √2 = a + b√2

Solution

(3 + √2) × (3 + √2) / (3 - √2) × (3 + √2)

= (3 + √2)^2 / (3)^2 - (√2)^2

= (3)^2 + (√2)^2 + 2 × 3 × √2 / 9 - 2

= 9 + 2 + 6√2 / 7

= 11 + 6√2/7

= 11/7 + 6√2/11

On getting the values

a = 11/7

b√2 = 6√2/11

b = 6/11





Thanks :))))
$hope \: it \: helps \: u$