Question

find the value of
$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } = a + b \sqrt{2}$
plzzzzz help fast...

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Answer 1

Answer:

a = $a =\frac{11}{7\\}, b =\frac{6}{7}$

Step-by-step explanation:

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

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

$\frac{(3+\sqrt{2})^{2} }{3^{2}-(\sqrt{2})^{2} } = a+b\sqrt{2}$       ( according to: $( a+b) (a-b) = a^{2} - b^{2}$ )

$\frac{3^{2}+ 2 . 3 .\sqrt{2} + (\sqrt{2})^{2} }{9-2} = a+b\sqrt{2}$

$\frac{9+6\sqrt{2}+2 }{7} = a+b\sqrt{2}$

$(11 + 6\sqrt{2}) 1 = 7(a + b\sqrt{2})$

$11+6\sqrt{2} = 7a+7b\sqrt{2}$

now,

$7a=11\\a = \frac{11}{7}$  

again,

$7b\sqrt{2} = 6\sqrt{2} \\7b =\frac{6\sqrt{2} }{\sqrt{2} }$           ( now divide $\sqrt{2}$ in $6\sqrt{2}$ by the denominator

$7b = 6\\b = \frac{6}{7}$