Answer
If a=0, then x=0
If a$\lt$0, then x=x=+2$\sqrt a$ i or x=-2$\sqrt a$ i or x=+$\sqrt a$ i or x=-$\sqrt a$ i
Work Step by Step
$x^{4}$-5a$x^{2}$+$4a^{2}$=0
Make $n^{2}$=$(x^{2})^{2}$=$x^{4}$
Then $n^{2}$-5an+4$a^{2}$=0
(n-a)(n-4a)=0
n=a or n=4a
Thus $x^{2}$=a or $x^{2}$=4a
If a$\gt$0, then x=+2$\sqrt a$ or x=-2$\sqrt a$ or x=+$\sqrt a$ or x=-$\sqrt a$
If a=0, then x=0
If a$\lt$0, then x=x=+2$\sqrt a$ i or x=-2$\sqrt a$ i or x=+$\sqrt a$ i or x=-$\sqrt a$ i
