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

1.	If x=root 2+1 ,find the value of x+1/x
Answer» $\mathfrak{\large{\underline{\underline{Answer:-}}}}$ $\boxed{ \tt x + \dfrac{1}{x} = 2 \sqrt{2} }$ $\mathfrak{\large{\underline{\underline{Explanation:-}}}}$ 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 $\tt x + \dfrac{1}{x} = \sqrt{2} + 1 + ( \sqrt{2} - 1)$ $\tt = \sqrt{2} + 1 + \sqrt{2} - 1$ $\tt = \sqrt{2} + \sqrt{2}$ $\tt = 2 \sqrt{2}$ $\Huge{\boxed{ \sf x + \dfrac{1}{x} = 2 \sqrt{2} }}$

Answer»

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