Question

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

Answer 1

$$\begin{gathered} \left(i\right) \\ \mathrm{Given},\left(x-\frac{1}{x}\right)=4 \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ {\left(x-\frac{1}{x}\right)}^2=4^2 \\ \Rightarrow {\left(x\right)}^2-2\times \left(x\right)\times \left(\frac{1}{x}\right)+{\left(\frac{1}{x}\right)}^2=16 \\ \Rightarrow x^2-2+\frac{1}{x^2}=16 \\ \Rightarrow x^2+\frac{1}{x^2}=16+2 \\ \therefore \left(x^2+\frac{1}{x^2}\right)=18 \end{gathered}$$² + 2 ⇒ 4² + 2 ⇒ 16 + 2 ⇒ 18

^{$$\begin{gathered} \left(ii\right) \\ \mathrm{From}\mathrm{the}\mathrm{first}\mathrm{part}: \\ \left(x^2+\frac{1}{x^2}\right)=18 \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ {\left(x^2+\frac{1}{x^2}\right)}^2={18}^2 \\ \Rightarrow \left(x^2\right){}^2+2\times \left(x^2\right)\times \left(\frac{1}{x^2}\right)+{\left(\frac{1}{x^2}\right)}^2=324 \\ \Rightarrow x^4+2+\frac{1}{x^4}=324 \\ \Rightarrow x^4+\frac{1}{x^4}=324-2 \\ \therefore x^4+\frac{1}{x^4}=322 \end{gathered}$$}