Question

Find the value of $x^2+\frac{1}{x^2}$, if $x+\frac{1}{x}=\sqrt{3}$

• A. $-1$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is B $1$

$x+\frac{1}{x}=\sqrt{3}$

We know that,

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

So,

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

${\left(\sqrt{3}\right)}^2=x^2+\frac{1}{x^2}+2$

$\Rightarrow x^2+\frac{1}{x^2}=3-2=1$