Question

please solve sum 8 part B .

Answer 1

$3 - 2 \sqrt{2}$

Answer 2

(8) - B:

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

On rationalizing the denominator, we get

$= \ \textgreater \ \frac{3 - 2 \sqrt{2} }{3 + 2 \sqrt{2} } * \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} }$

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

$= \ \textgreater \ \frac{(3 - 2 \sqrt{2} )^2}{(3)^2 - ( 2\sqrt{2})^2 }$

$= \ \textgreater \ \frac{9 + 8 - 12 \sqrt{2} }{9 - 8}$

$= \ \textgreater \ \frac{17 - 12 \sqrt{2} }{1}$

$= \ \textgreater \ 17 - 12 \sqrt{2}$

$\boxed {17 - 12 \sqrt{2} }$


Hope this helps!