দেওয়া আছে,
$2x+\frac2x=3$
বা, $2\left(x+\frac{1}{x}\right)=3$
বা, $\left(x+\frac{1}{x}\right)=\frac32$
বা, $\left(x+\frac{1}{x}\right)^2=\left(\frac32\right)^2$
বা, $x^2+2\cdot x \cdot \frac1x +\frac{1}{x^2}=\frac94$
বা, $x^2+2+\frac{1}{x^2}=\frac94$
বা, $x^2+\frac{1}{x^2}=\frac94-2$
বা, $x^2+\frac{1}{x^2}=\frac{9-8}{4}$
$\therefore x^2+\frac{1}{x^2}=\frac14$ [Answer]