If $a,b,c,r \in R$, then the equation $(x^2+ax-3b)(x^2-cx+b)(x^2-dx+2b)=0$ has

Given equation : $[x^2 + ax - 3b] [x^2 - cx + b][x^2 - dx + 2b] = 0$

For $x^2 + ax - 3b = 0$

Discriminant, $D = a^2 + 12 b$

For $x^2 - cx + b = 0$

Discriminant, $D = c^2 - 4b$

For $x^2 - dx + 2b = 0$

Discriminant, $D = d^2 + 8b$

Let us assume that $x^2 + ax - 3b$ has imaginary roots, then,

$a^2 + 12b < 0$

$b < -a^2 /12$

this implies b is negative, then $-4b$ will be positive

so,

$c^2 - 4b > 0$

$x^2 - cx + b = 0$ has real roots.

Therefore, $(x^2 + ax - 3b)(x^2 - cx + b)(x^2 - dx - 2b) = 0$ has at least 2 real roots.