Question

If $x+\frac{1}{x}=9,$ find the value of $x^4+\frac{1}{x^4}.$

Answer 1

Let us consider the following equation:

$x+\frac{1}{x}=9$

Squaring both sides, we get:

$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^2={\left(9\right)}^2=81 \\ \Rightarrow {\left(x+\frac{1}{x}\right)}^2=81 \\ \Rightarrow x^2+2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=81 \\ \Rightarrow x^2+2+\frac{1}{x^2}=81 \end{gathered}$$
$\Rightarrow x^2+\frac{1}{x^2}=79$ (Subtracting 2 from both sides)

Now, squaring both sides again, we get:

$$\begin{gathered} {\left(x^2+\frac{1}{x^2}\right)}^2={\left(79\right)}^2=6241 \\ \Rightarrow {\left(x^2+\frac{1}{x^2}\right)}^2=6241 \\ \Rightarrow {\left(x^2\right)}^2+2\left(x^2\right)\left(\frac{1}{x^2}\right)+{\left(\frac{1}{x^2}\right)}^2=6241 \\ \Rightarrow x^4+2+\frac{1}{x^4}=6241 \end{gathered}$$
$\Rightarrow x^4+\frac{1}{x^4}=6239$