Question

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

Answer 1

$x-\frac{1}{x}=3\text{Squaring on both sides,}\left(i\right){(x-\frac{1}{x})}^2=\left(3{)}^2\right(x{)}^2-2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=9x^2-2+\frac{1}{x^2}=9x+\frac{1}{x^2}=9+2=11x^2+\frac{1}{x^2}=11\left(ii\right)\text{Again squaring on both sides,}{(x^2+\frac{1}{x^2})}^2=\left(11{)}^2\right(x{)}^2\times 2\times x^2\times \frac{1}{x^2}+{\left(\frac{1}{x^2}\right)}^2=121x^4+2+\frac{1}{x^4}=121x^4+\frac{1}{x^4}=121-2=119x^2+\frac{1}{x^2}=11,x^4+\frac{1}{x^4}=119$