Question

If $x+\frac{1}{x}=2$ ; x $\ne$ 0, then value of $x^2+\frac{1}{x^2}$ is equal to:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

Given, $x+\frac{1}{x}=2--\left(i\right)$.

We know, $(a+b{)}^2=a^2+2ab+b^2$.

Then, on squaring $\left(i\right)$,

${\left(x+\frac{1}{x}\right)}^2=4$

$\Rightarrow x^2+\frac{1}{x^2}+2\times x\times \frac{1}{x}=4$

$\Rightarrow x^2+\frac{1}{x^2}+2=4$

$\Rightarrow x^2+\frac{1}{x^2}=4-2$

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

Therefore, the required answer is $2$.

Hence, option $B$ is correct.