Sketch the curve $\mathrm{f}(x)=x^{3}+Ax^{2}+B$ first in the case $A>0$ and $B>0$, and then in the case $A<0$ and $B>0.$
Show that the equation $x^{3}+ax^{2}+b=0$, where $a$ and $b$ are real, will have three distinct real roots if $27b^{2}+3a^{3}b<0,$ but will have fewer than three if $27b^{2}+4a^{3}b<0.$
I worked through this question. I give a sketch: $f'(x) = 3x^2 + 2Ax$ so the turning points are at 0 and $-\frac{2A}{3}$.
$f(-2A/3) = \frac{4A^3}{27} + B$
If $A>0$, $B\geq 0$, the both stationary points of the cub are non-negative. - 1 solution or 2
If $A<0$, $B\geq 0$, then we need $f(-2A/3)<0$ for 3 solutions, else there is only 1 or 2.
Similar argument can be made when $B<0$.
The condition I arrived was: depending on whether $27b^2+4a^3b<0$, we have 3 or less solutions. This is different to the result in the question. Why is that?
This is STEP Question 93-S1-Q7 found here: