Question

If x=2+√3, find the value of x+1/x

Answer 1

Hey mate!

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Given :

$x = 2 + \sqrt{3}$

To find :

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

Solution :

$x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{(2) {}^{2} - ( \sqrt{3}) {}^{2} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \frac{1}{x} = 2 - \sqrt{ 3}$

Now,

$x + \frac{1}{x} \\ \\ \implies 2 + \cancel{\sqrt{3}} + 2 - \cancel{ \sqrt{3}} \\ \\ \implies 2 + 2 \\ \\ \implies 4$

Hence,

The answer is 4.

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Thanks for the question !

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Answer 2

Answer:

4

Step-by-step explanation:

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

To Find : find the value of $x+\frac{1}{x}$

Solution:

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

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

To find $\frac{1}{x}$

$x=2+\sqrt{3}$

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

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

$\frac{1}{x}= \frac{2 -\sqrt{3}}{4 - 3}$

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

Substitute the values

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

$2+\sqrt{3}+2- \sqrt{ 3}$

$4$

Hence  the value of $x+\frac{1}{x}$ is 4