Question

The number of real roots of ${(x+\frac{1}{x})}^2-4=0$ is

• A. $0$
• B. $2$
• C. $4$
• D. none of these

Answer 1

The correct option is B $2$

${(x+\frac{1}{x})}^2-4=0{(x+\frac{1}{x})}^2-2^2=0(x+\frac{1}{x}-2)(x+\frac{1}{x}+2)=0\therefore x+\frac{1}{x}-2=0x^2-2x+1=0{(x-1)}^2=0x=1orx+\frac{1}{x}+2=0x^2+2x+1=0{(x+1)}^2=0\therefore x=-1\therefore 2realroots$