Question

If $x^2+\frac{1}{x^2}=38,$ then what is the value of $x-\frac{1}{x}$?

• A. $5$
• B. $6$
• C. $9$
• D. $4$

Answer 1

The correct option is B $6$

Let us consider the square of $x-\frac{1}{x}$ as follows:

${(x-\frac{1}{x})}^2=x^2+{\left(\frac{1}{x}\right)}^2-(2\times x\times \frac{1}{x})(\because (a-b{)}^2=a^2+b^2-2ab)=x^2+\frac{1}{x^2}-2=38-2(Given:x^2+\frac{1}{x^2}=38)=36$

Now, since ${(x-\frac{1}{x})}^2=36$, therefore, we have:

${(x-\frac{1}{x})}^2=36\Rightarrow x-\frac{1}{x}=\sqrt{36}=6$

Hence, $x-\frac{1}{x}=6$