Question

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

Answer 1

Answer:

1/x=√2+1 because x and 1/x are reciprocals of each other and the reciprocal of √2-1 is √2+1

here is the work...

1/x=1/(√2-1)

rationalize the denominator

1(√2+1)/(√2-1)(√2+1)

(√2+1)/(2-√2+√2-1)

(√2+1)/(2+0-1)

(√2+1)/1

√2+1

1/x=√2+1

Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

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

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

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

$\bf \underline{Solution-}$

$\textsf{We have}$

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

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

$\textsf{The denominator is √2 - 1.}$

$\textsf{We know that,}$

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

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

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

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

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

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

$\textsf{\: \: \: a = √2 and b = 1.}$

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

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

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

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

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

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

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

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

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

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

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