Question

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

Answer 1

$x+\frac{1}{x}=4$
Squaring both sides,
(i) ${(x+\frac{1}{x})}^2=(4{)}^2\Rightarrow x^2+\frac{1}{x^2}+2\times x+\frac{1}{x}=16\Rightarrow x^2+\frac{1}{x^2}+2=16\Rightarrow x^2+\frac{1}{x^2}=16-2=14\therefore x^2+\frac{1}{x^2}=14$

(ii) Again squaring both sides,
$(x^2+\frac{1}{x^2})=(14{)}^2\Rightarrow (x^2{)}^2+{\left(\frac{1}{x^2}\right)}^2+2\times x^2\times \frac{1}{x^2}=196\Rightarrow x^4+\frac{1}{x^4}+2=196\Rightarrow x^4+\frac{1}{x^4}=196-2=194\therefore x^{+}\frac{1}{x^4}=196-2=194\therefore x^4+\frac{1}{x^4}=194$