$\text{ The roots of the equation }\left(a^4+b^4\right)x^2+4abcdx+\left(c^4+d^4\right)=0$

The discriminant of the given equation is

$\begin{array}{l}\mathrm{D}=16{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2{\mathrm{d}}^2-4\left({\mathrm{a}}^4+{\mathrm{b}}^4\right)\left({\mathrm{c}}^4+{\mathrm{d}}^4\right)\\ =-4\left[\left({\mathrm{a}}^4+{\mathrm{b}}^4\right)\left({\mathrm{c}}^4+{\mathrm{d}}^4\right)-4{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2{\mathrm{d}}^2\right]\\ =-4\left[{\mathrm{a}}^4{\mathrm{c}}^4+{\mathrm{a}}^4{\mathrm{d}}^4+{\mathrm{b}}^4{\mathrm{c}}^4+{\mathrm{b}}^4{\mathrm{d}}^4-4{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2{\mathrm{d}}^2\right]\\ =-4\left[\left({\mathrm{a}}^4{\mathrm{c}}^4+{\mathrm{b}}^4{\mathrm{d}}^4-2{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2{\mathrm{d}}^2\right)+\left({\mathrm{a}}^4{\mathrm{d}}^4+{\mathrm{b}}^4{\mathrm{c}}^4-2{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2{\mathrm{d}}^2\right)\right]\\ =-4\left[{\left({\mathrm{a}}^2{\mathrm{c}}^2-{\mathrm{b}}^2{\mathrm{d}}^2\right)}^2+{\left({\mathrm{a}}^2{\mathrm{d}}^2-{\mathrm{b}}^2{\mathrm{c}}^2\right)}^2\right]----\left(1\right)\end{array}$

$\text{ If roots of the given equation are real, }D\ge 0$

$\begin{array}{l}\Rightarrow -4\left[{\left({\mathrm{a}}^2{\mathrm{c}}^2-{\mathrm{b}}^2{\mathrm{d}}^2\right)}^2+{\left({\mathrm{a}}^2{\mathrm{d}}^2-{\mathrm{b}}^2{\mathrm{c}}^2\right)}^2\right]\ge 0\\ \Rightarrow {\left({\mathrm{a}}^2{\mathrm{c}}^2-{\mathrm{b}}^2{\mathrm{d}}^2\right)}^2+{\left({\mathrm{a}}^2{\mathrm{d}}^2-{\mathrm{b}}^2{\mathrm{c}}^2\right)}^2\le 0\\ \Rightarrow {\left({\mathrm{a}}^2{\mathrm{c}}^2-{\mathrm{b}}^2{\mathrm{d}}^2\right)}^2+{\left({\mathrm{a}}^2{\mathrm{d}}^2-{\mathrm{b}}^2{\mathrm{c}}^2\right)}^2=0----\left(2\right)\end{array}$

['.' Sum of two positive quantities cannot be negative] 

From (1) and (2), we get D = 0 

Hence, the roots of the given quadratic equation are not different if they are real.