Question

a) x² − 5 |x| +1 = 0​

Answer 1

$x^2-5|x|+1=0\Leftrightarrow |x|^2-5|x|+1=0\\D_{|x|}=25-4=21\Rightarrow |x|=\cfrac{5\pm \sqrt{21}}{2}\Rightarrow x=\pm\cfrac{5\pm \sqrt{21}}{2}$