Question

If $x=\sqrt{5}+2$, then find the value of $x^2+\frac{1}{x^2}$ [ 4 Marks ]

Answer 1

$x=\sqrt{5}+2$[Given]
$\Rightarrow \frac{1}{x}=\frac{1}{\sqrt{5}+2}$ [ 1 Mark ]
$\Rightarrow \frac{1}{x}=\frac{[1\times (\sqrt{5}-2\left)\right]}{\left[\right(\sqrt{5}+2)\times (\sqrt{5}-2\left)\right]}$
$\Rightarrow \frac{1}{x}=\frac{\sqrt{5}-2}{(5-4)}$
$\Rightarrow \frac{1}{x}=\sqrt{5}-2$ [1 Mark ]
Now, $x^2+\frac{1}{x^2}=(x+\frac{1}{x}{)}^2-2$
$=(\sqrt{5}+2+\sqrt{5}-2{)}^2-2$
$=(2\sqrt{5}{)}^2-2$
$=20-2=18$
$\therefore x^2+\frac{1}{x^2}=18$ [ 2 Marks ]