Question

$\text{If}{x}^2+\frac{1}{x^2}=38,\text{find the values of:}$

$\left(i\right)x-\frac{1}{x}\left(ii\right)x^4+\frac{1}{x^4}$

Answer 1

$\left(i\right){(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2\times x\times \frac{1}{x}$

$[\because (a-b{)}^2=a^2+b^2-2ab]$

$x^2+\frac{1}{x^2}-2$

$\therefore x-\frac{1}{x}=\sqrt{36}=6$

$\left(ii\right){(x^2+\frac{1}{x^2})}^2$

$=x^4+\frac{1}{x^4}+2\times x^2\times \frac{1}{x^2}$

$\Rightarrow (38{)}^2=x^4+\frac{1}{x^4}+2$

$[\text{Given}(x^2+\frac{1}{x^2})=38]$

$\Rightarrow 1444-2=x^4+\frac{1}{x^4}$

$\Rightarrow 1442=x^4+\frac{1}{x^4}$

$\therefore x^4+\frac{1}{x^4}=1442$