Question

If $\left(x-\frac{1}{x}\right)=5$, find the values of
(i) $\left(x^2+\frac{1}{x^2}\right)$
(ii) $\left(x^4+\frac{1}{x^4}\right).$

Answer 1

$$\begin{gathered} \left(i\right)\left(x-\frac{1}{x}\right)=5 \\ \Rightarrow \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2={\left(5\right)}^2 \\ \Rightarrow \left(x^2+\frac{1}{x^2}-2\left(x\right)\left(\frac{1}{x}\right)\right)=25 \\ \Rightarrow \left(x^2+\frac{1}{x^2}\right)-2=25 \\ \Rightarrow \left(x^2+\frac{1}{x^2}\right)=25+2 \\ \Rightarrow \left(x^2+\frac{1}{x^2}\right)=27 \\ \mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{of}\left(x^2+\frac{1}{x^2}\right)\mathrm{is}27. \end{gathered}$$

$$\begin{gathered} \left(x^2+\frac{1}{x^2}\right)=27 \\ \Rightarrow \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ \Rightarrow \left(x^4+\frac{1}{x^4}-2\left(x^2\right)\left(\frac{1}{x^2}\right)\right)={\left(27\right)}^2 \\ \Rightarrow \left(x^4+\frac{1}{x^4}\right)-2=729 \\ \Rightarrow \left(x^4+\frac{1}{x^4}\right)=729+2 \\ \Rightarrow \left(x^4+\frac{1}{x^4}\right)=731 \\ \mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{of}\left(x^4+\frac{1}{x^4}\right)\mathrm{is}731. \end{gathered}$$