[Algebra]
Problem: If $x+\frac{1}{x}=\frac{13}{6}$, determine all values of $x^{2}+\frac{1}{x^{2}}$.

Solution: $\left(x+\frac{1}{x}\right)^{2}=\left(\frac{13}{6}\right)^{2}$; squaring

$x^{2}+2+\frac{1}{x^{2}}=\frac{169}{36}$

$x^{2}+\frac{1}{x^{2}}=\frac{169}{32}-2$

$x^{2}+\frac{1}{x^{2}}=\frac{169}{36}-\frac{72}{36}=\frac{97}{36}$
$6 x\left(x+\frac{1}{x}\right)=6 x\left(\frac{13}{6}\right)$

$6 x^{2}+6=13 x$

$6 x^{2}-13 x+6=0$

$(3 x-2)(2 x-3)=0$



$x=\frac{2}{3}$ or $x=\frac{3}{2}$

For $x=\frac{2}{3}, x^{2}+\frac{1}{x^{2}}$

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

$=\frac{4}{9}+\frac{9}{4}$

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

$=\frac{9}{4}+\frac{4}{9}$

$=\frac{97}{36}$

$=\frac{81+16}{36}$

$=\frac{97}{36}$

Final answer: $\frac{97}{36}$