Question

If $x^2+\frac{1}{x^2}=47$ then the value of $x+\frac{1}{x}$ is____

Answer 1

Answer: 7

$(x+\frac{1}{x}{)}^2=x^2+\frac{1}{x^2}+2\times x\times \frac{1}{x}$
$=x^2+\frac{1}{x^2}+2$
$=47+2$
$=49$
Thus,
$(x+\frac{1}{x}{)}^2=49$
$\Rightarrow x+\frac{1}{x}=\sqrt{49}=7$