Question

If $a=6-\sqrt{35}$ find the value of $a^2+\frac{1}{a^2}$

Answer 1

Given:$a=6-\sqrt{35}$

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

Now,$a^2+\frac{1}{a^2}={(a+\frac{1}{a})}^2-2a\times \frac{1}{a}$

or $a^2+\frac{1}{a^2}={(a+\frac{1}{a})}^2-2$

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

$={\left(12\right)}^2-2=144-2=142$

$\therefore a^2+\frac{1}{a^2}=142$