Question

if x= 1- underoot 2, find value of (x-1/x)to the power of 4

Answer 1

Very easy question just put the value nd get the answer

Answer 2

Hey friend, Harish here.

Here is your answer:

Given that,

x = 1 - √2 

To find,

$(x - \frac{1}{x})^{4}$

Solution,

We know that.

$x = 1- \sqrt{2}$

Then,

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

Now, Rationalize the number by multiplying and dividing the number by it's conjugate.

Conjugate of the number is 1 + √2.

Then,

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

We know that, (a+b)(a-b) = a² - b².

So. (1 + √2) (1 - √2) = 1² - (√2)² = 1 -2  = -1.

Then,

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

Now subtract, x & 1/x.

Then,

$(x- \frac{1}{x}) = [(1- \sqrt{2}) - (-1 - \sqrt{2})]$

→ $(x- \frac{1}{x}) = [1- \sqrt{2} +1 + \sqrt{2})]$

Here ( √2 & -√2  get cancelled)

→ $(x- \frac{1}{x}) = 1 + 1 = 2$

Now, Square both the sides then,

Then,

$(x- \frac{1}{x})^{2} = 2^{2} = 4$

Now again square both the sides,

Then.

$(x- \frac{1}{x})^{2\times 2} = 4^{2}$

→ $(x- \frac{1}{x})^{4} = 16.$

Therefore the value is 16.
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Hope my answer is helpful to you.