Question

if x = 1-√2 find the value of x-1/x

Answer 1

x=1-√2

x-1/x=1-√2-1/1-√⇒x-1/x=−  2  +3  +1≈−1.10100299

Answer 2

Given:

The value of x -

$\bf \to \: x = 1-\sqrt{2}$

What To Find:

We have to find the value of -

$\bf \to \: x - \dfrac{1}{x}$

How To Find:

To find we have to -

• First, we have to find the value of  $\sf \dfrac{1}{x}$.
• Next, we have to subtract $\sf x$ to $\sf \dfrac{1}{x}$.

• Then, we will get the answer.

Solution:

• Finding the value of -

$\sf \to \: \dfrac{1}{x}$

We know that -

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

Substitute the value in,

$\sf \to \: \dfrac{1}{x} = \dfrac{1}{1-\sqrt{2}}$

Rationalise the denominator with its rationalising factor,

$\sf \to \: \dfrac{1}{x} = \dfrac{1}{1-\sqrt{2}} \times \dfrac{1+\sqrt{2}}{1+\sqrt{2}}$

Take them as common,

$\sf \to \: \dfrac{1}{x} = \dfrac{1+\sqrt{2}}{(1-\sqrt{2}) \times (1+\sqrt{2})}$

Use the identity (a - b) (a + b) = a² - b²

$\sf \to \: \dfrac{1}{x} = \dfrac{1+\sqrt{2}}{(1)^2- (\sqrt{2})^2}$

Find the squares,

$\sf \to \: \dfrac{1}{x} = \dfrac{1+\sqrt{2}}{1- 2}$

Subtract the values,

$\sf \to \: \dfrac{1}{x} = \dfrac{1+\sqrt{2}}{-1}$

Can be written as,

$\sf \to \: \dfrac{1}{x} = -1+\sqrt{2}$

• Finding the value of -

$\sf \to \: x - \dfrac{1}{x}$

Substitute the values,

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

Remove the brackets,

$\sf \to \: 1-\sqrt{2} + 1-\sqrt{2}$

Rearrange the terms,

$\sf \to \: 1+1 - \sqrt{2} -\sqrt{2}$

Add the first terms,

$\sf \to \: 2 - \sqrt{2} -\sqrt{2}$

Subtract the second terms,

$\sf \to \: 2 - 2\sqrt{2}$

Final Answer:

∴ Thus, the answer is 2 - 2√2.