Every cubic equation must have atleast one real root, as because only two cases are possible for the roots of a cubic equation
- All 3 roots are real
- One real root and two non-real complex conjugate roots
Thus Option(B) is always correct.
Now,
For Option(A) -> Complex roots are also possible hence false
For Option(C) -> Since the coefficients $p,q$ and $r$ are all positive, $\alpha \beta \gamma =-r$ is negative (where $\alpha , \beta$ and $\gamma$ are the three roots of the cubic equation) , thus atleast one root has to be negative to make the product of the three roots negative.Hence false.
For Option (D) As stated above if all three roots are negative then also the product of three roots will be negative.Hence false.
For derving the formulae $\alpha \beta \gamma =-r$ , check Vieta's formulas.