Question 563771
x^4 + 5ax^2 + 4a^2 = 0
let u=x^2
u^2=x^4
u^2+5au+4a^2=0
(u+4a)(u+a)=0
u+4a=0
u=-4a
x^2=-4a
no real solution
or
u+a=0
u=-a
x^2=-a
Also, no real solution