Question

X=1-root2 find value of (x-1/x)*4

Answer 1

GIVEN:

★$x=1-\sqrt{2}$

TO FIND:

★$(x-\dfrac{1}{x}) ^{4}$

ANSWER:

We have,

$x=1-\sqrt{2}$

So,

=>$\dfrac{1}{x}=\dfrac{1}{1-\sqrt{2}}$

On rationalising denominator,

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

=>$\dfrac{1}{x}=\dfrac{1+\sqrt{2}}{1^{2}-\sqrt{2}^{2}}$

$\large\green{\boxed{(a+b) (a-b) =a^{2}-b^{2}}}$

=>$\dfrac{1}{x}=\dfrac{1+\sqrt{2}}{1-2}$

=>$\dfrac{1}{x}=\dfrac{1+\sqrt{2}}{-1}$

=>$\dfrac{1}{x}=\dfrac{(-1) (1+\sqrt{2})}{-1\times-1}$

=>$\dfrac{1}{x}=-1-\sqrt{2}$

______________________________________

Hence,

=>$(x-\dfrac{1}{x}) ^{4}=[1-\sqrt{2}-(-1-\sqrt{2})]^{4}$

=>$(x-\dfrac{1}{x}) ^{4}=[1-\sqrt{2}+1+\sqrt{2}]^{4}$

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

.°.$(x-\dfrac{1}{x}) ^{4}=16$

Hence required answer is 16.

$\huge\orange{\boxed{.°.(x-\dfrac{1}{x}) ^{4}=16}}$