Question

if $a=6-\sqrt{35},$ find the values of the following:
$\left(a\right)a^2+\frac{1}{a^2}$

$\left(b\right)a^2-\frac{1}{a^2}$

$\left(a\right)a^4-\frac{1}{a^4}$

Answer 1

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

Then $a^2=(6-\sqrt{35}{)}^2$

$=36-2\left(6\right)\left(\sqrt{35}\right)+35$

$=71-12\sqrt{35}$

Now $\frac{1}{a^2}=\frac{1}{71.12\sqrt{35}}=\frac{71+12\sqrt{35}}{(71{)}^2-(12\sqrt{35}{)}^2}=71+12\sqrt{35}$

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

and $a^2-\frac{1}{a^2}=-24\sqrt{35}$

Now $a^4=(71-12\sqrt{35}{)}^2=5,041-2\times 71\times 12\sqrt{35}+5040$

$10,081-1704\sqrt{35}$

$\frac{1}{a^4}=\frac{1}{10,081-1704\sqrt{35}}\times \frac{10081+704\sqrt{35}}{10081+1704\sqrt{35}}$

$=\frac{10081+1704\sqrt{35}}{1}$

$\Rightarrow a^4-\frac{1}{a^4}=-3403\sqrt{35}$