Question

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

Answer 1

$x-\frac{1}{x}=5$
(i) Squaring both sides
$(x-\frac{1}{x})=(5{)}^2$
$\Rightarrow x^2+\frac{1}{x^2}-2\times x\times \frac{1}{x}=25$
$\Rightarrow x^2+\frac{1}{x^2}-2=25$
$\Rightarrow x^2+\frac{1}{x^2}=25+2=27$
$\therefore x^2+\frac{1}{x^2}=27$

(ii) Again squaring both sides:
${(x^2+\frac{1}{x^2})}^2=(27{)}^2$
$\Rightarrow x^4+\frac{1}{x^4}+2\times x^2\times \frac{1}{x^2}=729$
$\Rightarrow x^4+\frac{1}{x^4}+2=729$
$\Rightarrow x^4+\frac{1}{x^4}=729-2=727$
$\therefore x^4+x^4+\frac{1}{x^4}=727$