If $p , q , r , s$ belongs to set of real numbers then equation $(x^2+px+3q)(-x^2+rx+q)(-x^2+sx-2q) = 0$: what type of roots does it have?

Question

It is an objective type question its answers are :

A - $6$ real roots

B - at least $2$ real roots

C - $2$ real and $4$ imaginary roots

D - $4$ real and $2$ imaginary roots

there could be multiple choices

Answer 1

We have no information regarding positivity, negativity or relative magnitudes, so it's impossible to determine how the roots will come.

A simple performance of the quadratic formula on each bracket shows us the three discriminants we are after. $$p^2-12q$$$$r^2+4q$$$$s^2-8q$$ whether these three are $<0$, or $>0$ is what we need to find.

The second one is only negative for negative $q$, in this case the other two are positive. This implies $4$ real roots, and thereby two complex. However, should $p^2<12q$ and $s^2<8q$, with $q>0$, then the second is positive but the other two are negative: four complex solutions and two reals. Moreover, if $p^2>12q$ and $s^2>8q$, with $q>0$, all three discriminants are positive and so we have six real solutions.

So the correct answer is $(B)$ there are at least two real solutions.