Question

If $x-\frac{1}{x}=3,$ find the values of $x^2+\frac{1}{x^2}$ and $x^4+\frac{1}{x^4}.$

Answer 1

Let us consider the following equation:

$x-\frac{1}{x}=3$

Squaring both sides, we get:

$$\begin{gathered} {\left(x-\frac{1}{x}\right)}^2={\left(3\right)}^2=9 \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2=9 \\ \Rightarrow x^2-2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=9 \\ \Rightarrow x^2-2+\frac{1}{x^2}=9 \end{gathered}$$
$\Rightarrow x^2+\frac{1}{x^2}=11$ (Adding 2 to both sides)

Squaring both sides again, we get:

$$\begin{gathered} {\left(x^2+\frac{1}{x^2}\right)}^2={\left(11\right)}^2=121 \\ \Rightarrow {\left(x^2+\frac{1}{x^2}\right)}^2=121 \\ \Rightarrow {\left(x^2\right)}^2+2\left(x^2\right)\left(\frac{1}{x^2}\right)+{\left(\frac{1}{x^2}\right)}^2=121 \\ \Rightarrow x^4+2+\frac{1}{x^4}=121 \end{gathered}$$
$\Rightarrow x^4+\frac{1}{x^4}=119$