Question

If $x=6-\sqrt{35},$ then find the value of $x^2+\frac{1}{x^2}.$

Answer 1

$x=6-\sqrt{35}$

$\frac{1}{x}=\frac{1}{6-\sqrt{35}}\times \frac{6+\sqrt{35}}{6+\sqrt{35}}$

$=\frac{6+\sqrt{35}}{36-35}=\frac{6+\sqrt{35}}{1}=6+\sqrt{35}$

$x^2+\frac{1}{x^2}={(6-\sqrt{35})}^2+{(6+\sqrt{35})}^2$

$=36+{\left(\sqrt{35}\right)}^2-12\sqrt{35}+36+{\left(\sqrt{35}\right)}^2+12\sqrt{35}$

$=2(36+35)$

$=2\times 71=142$