The roots of the equation $2a^2x^2 ;- 2abx +b^2=0$ when a < 0 and b > 0 are :& ;

For an equation $ax^2 \ + \ bx \ +c =0$, where $a \neq \ 0$ to have real roots, D must be greater than or equal to 0, where

$D = b^2-4ac$

In the given equation, $2a^2 x^2 \ - \ 2abx \ + \ b^2 =0$

$D=(-2ab)^2-4\times(2a^2)\times(b^2)$

$\Rightarrow D= -6 a^2 b^2$

$\because a,b \neq 0 \ \Rightarrow -6a^2 b^2$ is always negative

We see that $D$ is strictly negative

Hence, the equation always has complex roots.