Question

If $x^2+\frac{1}{x^2}=23$ find the values of $(x+\frac{1}{x}),(x-\frac{1}{x})and(x^4+\frac{1}{x^4})$

Answer 1

$x^2+\frac{1}{x^2}=23$
$\Rightarrow x^2+\frac{1}{x^2}+2=25$ [Adding 2 on both sides of (i)]
$\Rightarrow \left(x^2\right)+{\left(\frac{1}{x}\right)}^2+2.x.\frac{1}{x}=25$
$\Rightarrow {(x+\frac{1}{x})}^2=(5{)}^2$
$\Rightarrow x+\frac{1}{x}=\pm 5$
$\Rightarrow {(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2$
$\Rightarrow {(x-\frac{1}{x})}^2=23-2=21$
$\Rightarrow (x-\frac{1}{x})=\pm \sqrt{21}$
${(x^2+\frac{1}{x^2})}^2=(x^4+\frac{1}{x^4})+2$
$\Rightarrow (x^4+\frac{1}{x^4})={(x^2+\frac{1}{x^2})}^2-2$
$\Rightarrow (x^4+\frac{1}{x^4})=(23{)}^2-2=529-2$
$\Rightarrow (x^4+\frac{1}{x^4})=527$