Question

find the value of A and B : - $3 +\sqrt{2} /3- \sqrt{2} = A + B\sqrt{2}$

Answer 1

Given that,

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

To find the values of a and b

Now,

$\sf{ \frac{3 + \sqrt{2} }{3 - \sqrt{2} } } \\ \\ = \frac{3 + \sqrt{2} }{3 - \sqrt{2} } \times \frac{3 + \sqrt{2} }{3 - \sqrt{2} } \\ \\ = \frac{(3 + \sqrt{2) {}^{2} } }{3 {}^{2} - ( \sqrt{2}) {}^{2} } \\ \\ = 9 + 2 + 6 \sqrt{2} \\ \\ = \huge{11 + 6 \sqrt{2}}$

On comparing,

a=11 and b=6

Answer 2

$3 + 1 \div 3 - 1 = a + b(1)$

$- 1 = a + b$

HENCE, A = (-0.5)

B = (-0.5)