A student answered the following question: $a$ and $b$ are non-zero real numbers. Prove that the equation $x ^ { 3 } + a x ^ { 2 } + b = 0$ has three distinct real roots if $27 b \left( b + \frac { 4 a ^ { 3 } } { 27 } \right) < 0$
Here is the student's solution: I We differentiate $y = x ^ { 3 } + a x ^ { 2 } + b$ to get $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } + 2 a x = x ( 3 x + 2 a )$ Solving $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ shows that the stationary points are at $( 0 , b )$ and $\left( - \frac { 2 a } { 3 } , b + \frac { 4 a ^ { 3 } } { 27 } \right)$
II If $27 b \left( b + \frac { 4 a ^ { 3 } } { 27 } \right) < 0$, then $b$ and $b + \frac { 4 a ^ { 3 } } { 27 }$ must have opposite signs, and so one of the stationary points is above the $x$-axis and one is below.
III If the cubic has three distinct real roots, then one of the stationary points is above the $x$-axis and one is below.
IV Hence if $27 b \left( b + \frac { 4 a ^ { 3 } } { 27 } \right) < 0$, then the equation has three distinct real roots.
Which one of the following options best describes the student's solution?
A It is a completely correct solution.
B The student has instead proved the converse of the statement in the question.
C The solution is wrong, because the student should have stated step II after step III.
D The solution is wrong, because the student should have shown the converse of the result in step II.
E The solution is wrong, because the student should have shown the converse of the result in step III.