Question

By Rationalizing the denominator of 1/(3+√2) we get
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Answer 1

Answer:

Given problem is \frac{1}{3-2 \sqrt{2}}

3−2

2

1

To rationalize, we multiply by conjugate of the denominator.

Conjugate is obtained by changing the sign of radical term

So conjugate of 3-2 \sqrt{2}3−2

2

is 3+2 \sqrt{2}3+2

2

Hence we multiply and divide by 3+2 \sqrt{2}3+2

2

=\frac{1}{3-2 \sqrt{2}}* \frac{3+2 \sqrt{2}}{3+2 \sqrt{2}}=

3−2

2

1

∗

3+2

2

3+2

2

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

(3−2

2

)(3+2

2

)

3+2

2

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

(3)

2

−(2

2

)

2

3+2

2

= \frac{3+2 \sqrt{2}}{9-(4 *2)}=

9−(4∗2)

3+2

2

= \frac{3+2 \sqrt{2}}{9-8}=

9−8

3+2

2

= \frac{3+2 \sqrt{2}}{1}=

1

3+2

2

= 3+2 \sqrt{2}=3+2

2

Hence final answer is = 3+2 \sqrt{2}=3+2

2

.