Question

X=3+2√2 then find x+1/x

Answer 1

Hey friend,
Here is the answer you were looking for:
$x = 3 + 2 \sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \\ \\ on \: rationalizing \: we \: get \\ \\ \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ = \frac{3 - 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2}) }^{2} } \\ \\ = \frac{3 - 2 \sqrt{2} }{9 - 4 \times 2} \\ \\ = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \\ \frac{1}{x} = 3 - 2 \sqrt{2} \\ \\ x + \frac{1}{x} = (3 + 2 \sqrt{2} ) + (3 - 2 \sqrt{2} ) \\ \\ = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} \\ \\ = 3 + 3 + 2 \sqrt{2} - 2 \sqrt{2} \\ \\ = 3 + 3 \: \: \: ( + 2 \sqrt{2} \: and \: - 2 \sqrt{2} got \: cancelled) \\ \\ = 6$

Hope this helps!!!

@Mahak24

Thanks...
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Answer 2

Hey friend, Harish here.

Here is your answer:

Given that, 

x = 3 + 2√2.

To find,

$x + \frac{1}{x}$

Solution:

We know that,

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

Then,

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

Now, we must rationalize it's value.

So, Multiply and divide the number by it's conjugate.

The conjugate is 3 - 2√2.

Then,

 $\frac{1}{x} = \frac{1}{3+2 \sqrt{2} }\times \frac{3-2 \sqrt{2} }{3-2 \sqrt{2} }$

We know that (a +b) (a-b) = a² - b².

Then, (3+2√2) (3-2√2) = 3² - (2√2)² = 9 - 8 = 1.

So,

$\frac{1}{x}= \frac{3-2 \sqrt{2} }{(3+2 \sqrt{2})(3-2 \sqrt{2}) } = \frac{3-2 \sqrt{2}}{1} = 3-2 \sqrt{2}$

Now add both x & 1/x.

Then,

$x + \frac{1}{x}= [(3+2 \sqrt{2} ) + (3-2 \sqrt{2})]$

→ $x + \frac{1}{x}=(3+3)$

→ $x + \frac{1}{x}= 6$

Therefore the value is the 6.
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Hope my answer is helpful to you.