Question

if x = root2 - 1, find the value of x+1/x

Answer 1

Step-by-step explanation:

The value of x is given to be : x = √2 + 1

Now, we need to find the value of the given expression :

x+\frac{1}{x}x+x1

The value of the required expression can be found easily by substituting the value of x = √2 + 1 in the expression

So, The value is :

\begin{gathered}=x+\frac{1}{x}\\\\=(\sqrt{2}+1)+\frac{1}{(\sqrt{2}+1)}\\\\= \frac{(\sqrt{2}+1)^2+1}{\sqrt{2}+1}\\\\ =\frac{2+1+2 \sqrt{2}+1}{\sqrt{2}+1}\\\\=\frac{4+2\sqrt{2}}{\sqrt{2}+1}\\\\=2.83\end{gathered}=x+x1=(2+1)+(2+1)1=2+1(2+1)2+1=2+12+1+22+1=2+14+22=2.83

Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

$\sf{x = \sqrt{2} - 1}$

$\bf \underline{To\: find-}$

$\sf{the \: value \: of \: x + \frac{1}{x} = \: ?}$

$\bf \underline{Solution-}$

$\textsf{We have}$

$\sf{x = \sqrt{2} - 1}$

$\sf{ \therefore \: \frac{1}{x} = \frac{1}{ \sqrt{2} - 1 } }$

$\textsf{The denominator is √2 - 1.}$

$\textsf{We know that,}$

$\textsf{Rationalising factor of √a - b = √a + b. }$

$\textsf{So, the rationalising factor of √2 - 1 = √2 + 1.}$

$\textsf{So, rationalising the denominator them}$

$\sf{ \: \: = \frac{1}{ \sqrt{2} - 1} \times \frac{ \sqrt{2} + 1 }{ \sqrt{2} + 1} }$

$\sf{ \: \: = \frac{1( \sqrt{2} + 1)}{( \sqrt{2} - 1)( \sqrt{2} + 1) } }$

$\textsf{Now, comparing the denominator with (a-b)(a+b), we get}$

$\textsf{\: \: \: a = √2 and b = 1.}$

$\textsf{Using identity (a-b)+a+b) = a²-b², we have}$

$\sf{ \: \: = \frac{ \sqrt{2} + 1 }{( \sqrt{2} {)}^{2} - (1 {)}^{2} } }$

$\sf{ \: \: = \frac{ \sqrt{2} + 1}{2 - 1} }$

$\sf{ \: \: = \frac{ \sqrt{2} + 1}{1} }$

$\sf{ \: \: \frac{1}{x} = \sqrt{2} + 1 }$

$\textsf{Now, adding both values x and 1/x, we get}$

$\sf \therefore \: x + \frac{1}{x} = (\sqrt{2} - 1 )+ ( \sqrt{2} + 1 )$

$\sf \: x + \frac{1}{x} = \sqrt{2} \cancel{ - 1} + \sqrt{2} \cancel{+ 1 }$

$\sf \: x + \frac{1}{x} = \sqrt{2} + \sqrt{2}$

$\sf \: x + \frac{1}{x} =2 \sqrt{2}$

$\bf \underline{Hence, the\: value\: of \: x + \frac{1}{x} \: is \:2 \sqrt{2} . }$