Question

find a and b if 3+√2/3-√2=a+b√2​

Answer 1

Answer:

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

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

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

$a = 3 \: \: and \: b = - \frac{2}{3}$

Hope this answer helps you ^_^ !

Answer 2

Answer:

a=11/7 and b=6/7

Step-by-step explanation:

$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } = a + b \sqrt{2} \\ \frac{3 + \sqrt{2} }{3 - \sqrt{2} } \times \frac{3 + \sqrt{2} }{3 + \sqrt{2} } = a + b \sqrt{2} \\ \frac{(3 + \sqrt{2}) ^{2} }{({3})^{2} - ( \sqrt{2}) ^{2} } = a + b \sqrt{2} \\ \frac{9 + 2 + 6 \sqrt{2} }{9 - 2} = a + b \sqrt{2} \\ \frac{11 + 6 \sqrt{2} }{7} = a + b \sqrt{2} \\ \frac{11}{7} + \frac{6 \sqrt{2} }{7} = a + b \sqrt{2}\\ comparing \: both \: sides \: \: we \: get \\ a = \frac{11}{7} \\ b = \frac{6}{7}$