Real roots and variation of the constant term for a cubic polynomial

Let us define $f(x)=x^3+x^2$ . The zeros of its derivative $f'(x)=3x^2+2x$ are $x=-\frac{2}{3}$ and $x=0$ . One can also see that $f''(-\frac{2}{3})=-2$ and $f''(0)=2$ , so $f$ has a local maximum value of $\frac{4}{27}$ at $x=-\frac{2}{3}$ and a local minimum value of $0$ at $x=0$ .

In addition to that, the given problem can be reduced to find the interval of all values of $c$ such that the equation $f(x)=-c$ has two or three real solutions.

We are going to prove first that if the equation $f(x)=-c$ has three or two different real solutions then $-\frac{4}{27}\leq c\leq 0.$

Case 1 When the equation $f(x)=-c$ has three different real solutions: $x_{1}, x_{2}, x_{3}$ where $x_{1}< x_{2}<x_{3}$ .

Using the Rolle'sTheorem , we obtain that the smaller zero of $f'(x)$ , $x=-\frac{2}{3}$ must be in between $x_{1}$ and the $x_{2}$ and the larger zero, $x=0$ is in between $x_{2}$ and the $x_{3}$ , and it is easy to see the the local maximum value of $f(x)$ is also the absolute maximum value of $f(x)$ on the interval $[x_{1}, x_{2}]$ and the local minimum value of $f(x)$ is also the absolute minimum value on $[x_{2}, x_{3}]$ . Therefore, $0<-c<\frac{4}{27}$ .

Case 2 When the equation $f(x)=-c$ has two different real solutions: $x_{1}, x_{2}$ .

If the equation has only two different real solutions, then the third solution can't be complex, because the complex solutions appear in pairs of conjugate complex numbers (see Polynomial Roots ). Therefore one of the two real solutions is a root of multiplicity 2 of $P(x)=f(x)+c$ . It is not difficult to see that a root of multiplicity higher than one of a polynomial $P(x)$ is also a root of its derivative $P'(x)=f'(x).$ This implies that one of the two roots $x_{1}$ or $x_{2}$ is equal to either $-\frac{2}{3}$ or $0$ . Therefore, $-c=\frac{4}{27}$ or $-c=0$ .

From the conclusions of these two cases we get that when the given equation $f(x)=-c$ has more than one solution then we obtain that $0\leq-c\leq\frac{4}{27}$ that implies $-\frac{4}{27} \leq c\leq 0.$

Now we have to prove the converse implication. That is, if $-\frac{4}{27}\leq c\leq 0$ then the equation $f(x)=-c$ has more than one real solution.

If $c=-\frac{4}{27}$ , then $-c=\frac{4}{27}$ and one of the solutions of $f(x)=-c$ is $-\frac{2}{3}$ and we can prove that there is one more solution that is going to be in the interval $(0,1)$ by using the Intermediate Value Theorem because $f(0)+c=-\frac{4}{27}<0$ , and $f(1)=2-\frac{4}{27}>0$ .

If $c=0$ , then $-c=0$ and then one of the solutions of $f(x)=-c$ is $0$ and we can see that there is one more solution that is indeed $-1.$

If $-\frac{4}{27}< c < 0,$ then $f(-1)+c=c<0,$ $f(-\frac{2}{3})+c=\frac{4}{27}+c>0,$ $f(0)+c=c<0,$ and $f(1)+c=2+c>0$ . Therefore, using the Intermediate Value Theorem again we get that the equation has a solution in each of the following intervals: $(-1, -\frac{2}{3}),$ $(-\frac{2}{3}, 0),$ and $(0, 1).$ So the equation $f(x)=-c$ has 3 different solutions.

Now we can conclude that to get the answer to the problem we have to find the length of the interval $[-\frac{4}{27}, 0]$ which is $\frac{4}{27}$ where $4$ and $27$ are co-primes. Therefore the answer must be $4+27=31.$

Use calculus to see that the function $x^3+x^2$ has the local minimum $f(0)=0$ and the local maximum $f(-2/3)=4/27$ . Thus $P(x)=x^3+x^2+c$ has more than one real root for $-\frac{4}{27}\leq{c}\leq{0}$ , with $a+b=4+27=\boxed{31}$

Nice problem! I will assign it in a calculus exam one of these days ;)

Consider g(x)=-(x^3+x^2) and plot its graph.Rest is easy to follow.