Question

If x is 2 + root 3 then the value of x + 1/x is

Answer 1

Heya friend,
Here is the answer you were looking for:
$x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} }$

On rationalizing the denominator we get,

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

Using the identity :

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

$\frac{1}{x} = \frac{2 - \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3} )}^{2} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \frac{1}{x} = 2 - \sqrt{3} \\ \\ = x + \frac{1}{x}$
Putting the values :


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


Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
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Answer 2

$\huge\underline\mathfrak\orange{Answer-}$

value of $x+\dfrac{1}{x}$ = 4

$\huge\underline\mathfrak\orange{Explanation-}$

Given :

• x = 2 + √3

To find :

• Value of x + $\dfrac{1}{x}$

Solution :

It is given that,

x = 2 + √3

We have to find the value of $\dfrac{1}{x}$

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

Rationalising the denominator,

$\implies$ $\dfrac{1}{2+\sqrt{3}}$ × $\dfrac{2-\sqrt{3}}{2-\sqrt{3}}$

$\implies$ $\dfrac{2-\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}$

We know that,

(a+b)(a-b) = a² - b²

$\implies$ $\dfrac{2-\sqrt{3}}{{2}^{2}-3}$

$\implies$ $\dfrac{2-\sqrt{3}}{4-3}$

$\implies$ 2 - √3

Therefore, $\dfrac{1}{x}$ = 2 - √3

________________

Now, we have to find the value of x + $\dfrac{1}{x}$

$\implies$ ( 2 + √3 ) + ( 2 - √3 )

$\implies$ 2 + √3 + 2 - √3

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

$\implies$ 2 + 2

$\implies$ 4

Therefore, value of $\bold{x+\dfrac{1}{x}}$ = 4