Mathematics - Quadratic Equation Question with Solution | TestHub

If x is real, $x^2-6|x|+8=0\Rightarrow |x{|}^2-6|x|+8=0\Rightarrow |x|=2,4\Rightarrow x=\pm 2,\pm 4$ (4 real roots)

If x is non - real, say $x=\alpha +i\beta$, then $(\alpha +i\beta {)}^2-6\sqrt{{\alpha }^2+{\beta }^2}+8=0\left(|\alpha +i\beta |=\sqrt{{\alpha }^2+{\beta }^2}\right)$ $\left({\alpha }^2-{\beta }^2+8-6\sqrt{{\alpha }^2+{\beta }^2}\right)+2i\alpha \beta =0$

Comparing real and imaginary parts,

$\alpha \beta =0\Rightarrow \alpha =0$ (if $\beta =0$ then x is real.)

$\&-{\beta }^2+8-6\sqrt{{\beta }^2}=0$

${\beta }^2\pm 6\beta -8=0\Rightarrow \beta =\frac{\mp 6\pm \sqrt{68}}{2}$

ie., $\beta =\pm (3-\sqrt{17})$

Hence $\pm (3-\sqrt{17})i$ are non-real roots. (2 non real roots)