Question

If x=root 2+1 ,find the value of x+1/x

Answer 1

Step-by-step explanation:

√2+1+(1/√2+1) (taking LCM)

(2+1+2√2+1)/(√2+1)

(4+2√2)/√2+1

2√2(√2+1)/(√2+1)

=2√2

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Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\boxed{ \tt x + \dfrac{1}{x} = 2 \sqrt{2} }$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given :- $\tt x = \sqrt{2} + 1$

To find :- $\tt x + \dfrac{1}{x}$

Solution :-

First find the value of 1/x

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

Rationalise the denominator

The rationalising factor of √2 + 1 is √2 - 1. So multiply both numerator and denominator with rationalising factor

$\tt = \dfrac{1}{ \sqrt{2} + 1} \times \dfrac{ \sqrt{2} - 1}{ \sqrt{2} - 1}$

$\tt = \dfrac{ \sqrt{2} - 1}{( \sqrt{2} + 1)( \sqrt{2} -1)}$

$\tt = \dfrac{ \sqrt{2} - 1}{ {( \sqrt{2})}^{2} - {1}^{2} }$

[Since (x + y)(x - y) = x² - y² and here x = √2 and y = 1]

$\tt = \dfrac{ \sqrt{2} - 1}{2 - 1}$

$\tt = \dfrac{ \sqrt{2} - 1}{1}$

$\tt = \sqrt{2} - 1$

So we got value of 1/x as √2 - 1

Now we know values as $\bf x = \sqrt{2} + 1, \: \dfrac{1}{x} = \sqrt{2} - 1$

So now we can find x + 1/x

$\tt x + \dfrac{1}{x} = \sqrt{2} + 1 + ( \sqrt{2} - 1)$

$\tt = \sqrt{2} + 1 + \sqrt{2} - 1$

$\tt = \sqrt{2} + \sqrt{2}$

$\tt = 2 \sqrt{2}$

$\Huge{\boxed{ \sf x + \dfrac{1}{x} = 2 \sqrt{2} }}$