Question

If x2 +1/x2 =2 then x2-1/x2 = ?

Answer 1

Answer:

$\large\bold\red{0}$

Step-by-step explanation:

Given,

${x}^{2} + \frac{1}{ {x}^{2} } = 2 \\ = > {(x)}^{2} + {( \frac{1}{x} )}^{2} = 2$

Now,

We know that,

${a}^{2} + {b}^{2} = {(a - b)}^{2} + 2ab$

Therefore,

We get,

$= > {(x - \frac{1}{x} )}^{2} + 2 \times x \times \frac{1}{x} = 2 \\ = > {(x - \frac{1}{x}) }^{2} + 2 = 2 \\ = > {(x - \frac{1}{x} )}^{2} = 2 - 2 \\ = > {(x - \frac{1}{x}) }^{2} = 0 \\ = > x - \frac{1}{x} = 0 \\ = > x = \frac{1}{x} \: \: \: \: \: ........(i)$

Now,

We have to find,

${x}^{2} - \frac{1}{ {x}^{2} }$

But from Equation $(i)$,

We have,

$x = \frac{1}{x}$

Therefore,

Putting the value,

We get,

$= {( \frac{1}{x} )}^{2} - \frac{1}{ {x}^{2} } \\ = \frac{1}{ {x}^{2} } - \frac{1}{ {x}^{2} } \\ = 0$

Hence,

O is the required answer.

Answer 2

Answer: 0

Step-by-step explanation:

Given,

$x^2+\frac{1}{x^2}=2$

We know that by using,

(a - b)² = a² + b² + 2ab

We have,

$(x-\frac{1}{x})^2=x^2+\frac{1}{x^2}-2\times x\times\frac{1}{x}\\\;\\(x-\frac{1}{x})^2=2-2\\\;\\(x-\frac{1}{x})^2=0\\\;\\x-\frac{1}{x}=0$

Now,

$(x^2-\frac{1}{x^2})=(x+\frac{1}{x})(x-\frac{1}{x})\;\;\;\;\;\;\;\;\;[\because a^2-b^2=(a+b)(a-b)]$

$(x^2-\frac{1}{x^2})=(x+\frac{1}{x})\times 0\\\;\\(x^2-\frac{1}{x^2})=0$