On real roots of a polynomial equation

Question

Let $f(x)=x^3 + ax^2 + bx + c$ where $a,b,c \in \mathbb R$ such that if $f(r)=0$ , then $ f ' (r) \ne 0$ i.e. $f$ has no double-root in $\mathbb C$ i.e. $f$ has three distinct roots with at least one real root. Let $g(x) = 2 f''(x) f(x) - (f'(x))^2$.

Then how to show that $g$ has exactly two real roots ? If $r< s$ are the two real roots of $g(x)$, then how to show that $f(r) <0$ and $f(s) >0$ ?

Since $g'(x)=12 f(x)$, so $g$ has degree $4$ and since $f$ has no double root , so all the 4 roots of $g$ are also distinct. I am unable to say anything else. Please help.

Answer 1

As $g'(x)=12f(x)$, all local extrema of $g$ are at roots $r$ of $f$, $f(r)=0$. As these local extrema of $g$ the value $g(r)=-f'(r)^2<0$ is negative, even if $f$ has 3 real roots, the local maximum still has a negative value. As the leading term of $g$ is $3x^4$, for large $|x|$ the value of $g$ becomes positive. Thus there are roots of $g$ left and right of the root set of $f$. As $g$ is monotonous on those segments, there is exactly one root of $g$ left of the leftmost root of $f$ and one right of the rightmost one.

In other words, let $a=\min\{x\in\Bbb R:f(x)=0\}$ and $b=\max\{x\in\Bbb R:f(x)=0\}$. Then $g$ is negative on $[a,b]$ and monotonous falling resp. increasing on $(-\infty,a]$ and $[b,\infty)$ with a sign change and thus exactly one root in each of the intervals.

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Or another way using more directly the degree of $g$: if $g$ had $4$ real roots $s_1\le s_2\le s_3\le s_4$, then $g(x)=3(x-s_1)(x-s_2)(x-s_3)(x-s_4)$ would take non-negative values in the interval $[s_2,s_3]$, thus also at the local maximum $r$ there, which is impossible by the first observation, $g(r)=-f'(t)^2$ and $f'(r)\ne0$ as $r$ is also one root of $f$, and a simple one at that.

Answer 2

Updated 20.02

Using expressions for $f(x)$ and $g(x),$ one can get: $$f'(x) = 3x^2 + 2ax +b = 3\left(\left(x + \frac a3\right)^2 + d\right),\tag1$$ $$g(x) = 3x^4 + 4ax^3 + 6bx^2 + 12cx + 4ac - b^2,$$ $$g(x) = 3\left(\left(x + \frac a3\right)^2 + d\right)^2 - 36d^2 + 12k\left(x + \frac a3\right),\tag2$$ $$f(x) = \frac1{12}g'(x) = \left(x + \frac a3\right)\left(\left(x + \frac a3\right)^2 + d\right) + k,\tag3$$ where $$d = \frac b3 - \frac {a^2}9,\quad k = c - ad - \frac{a^3}{27}.$$

Case $\underline {d=0}.$

Taking in account $(1)$ and $(3),$ the issue condition $$f(x) = 0 \rightarrow f'(x) \not= 0\tag4$$ means that $k \not= 0.$ So $$g(x) = 3\left(x + \frac a3\right)\left(\left(x + \frac a3\right)^3 + 4k\right)$$ has exactly 2 real roots.

Case $\underline {d \not= 0}.$

Let $$y = x + \dfrac a3,$$ $$G(y) = g(y - \frac a3) = (3y^2 + d)^2 -36d^2 + 12 ky.$$ Easy to see that: $$G(0) = -27d^2 < 0,$$ $$G\left(\pm\infty\right) = +\infty.$$ So continuous function $G(x)$ has at least 2 real roots.

On the other hand, $$G''(y) = 12y(3y^2 + d).$$ So G(y) can't have more than one zero, one positive and one negative extremums. That means that it can't have more than two positive and two negative roots.

At the same time, according to Viete's theorem, function $G(y)$ has negative production of roots, so both of additional roots must have the same sign.

This contradiction means that both $G(x)$ and $g(x)$ have exactly 2 real roots in this case too.

We have shown that $g(x)$ has exactly 2 real roots.

Answer 3

As shown in LutzL's partial answer, $g$ must have at least two distinct real roots $\alpha,\beta$. Suppose by contradiction that $\frac{g(x)}{(x-\alpha)(x-\beta)}$ has a real root, which will then be a root of $g$ also. Then, since $g$ has degree $4$, all the roots of $g$ are real. Denote them by $g_1,g_2,g_3,g_4$. From $g(x)=3x^4 + 4ax^3 + 6bx^2 + 12cx + 4ac - b^2=3(x-g_1)(x-g_2)(x-g_3)(x-g_4)$, we deduce

$$ \begin{array}{lcl} a &=& -\frac{3}{4}\big(g_1+g_2+g_3+g_4\big), \\ b &=& \frac{1}{2}\big(g_1g_2+g_1g_3+g_2g_3+g_1g_4+g_2g_4+g_3g_4\big), \\ c &=& -\frac{1}{4}\big(g_1g_2g_3+g_1g_2g_4+g_1g_3g_4+g_2g_3g_4\big) \\ \end{array} \tag{1} $$

and the last coefficient of $g$ being $4ac-b^2=3g_1g_2g_3g_4$, we see that $G=0$ where

$$ G=\sum_{i,j} g_i^2g_j^2 +6g_1g_2g_3g_4 - \sum_{i,j,k} g_ig_jg_k^2 \tag{2} $$

Now, if we put $k_1=g_1^2+g_2^2+g_3^2-g_1g_2-g_1g_3-g_2g_3= \frac{1}{2}.((g_1-g_2)^2+(g_1-g_3)^2+(g_2-g_3)^2)$, $k_2=\sum_{i,j}g_i^2g_j-6g_1g_2g_3$ and $k_3=((g_1-g_2)(g_1-g_3)(g_2-g_3))$, we have the identity

$$ 16k_1G=(2k_1g_4+k_2)^2+3k_3^2 \tag{3} $$

It follows that $2k_1g_4+k_2=0$ and $k_3=0$. So at least two among $g_1,g_2,g_3$ are equal ; assume without loss that $g_2=g_1$. Then, $2k_1g_4+k_2=0$ becomes $2(g_3-g_1)^2(g_4-g_1)=0$, which forces $g_1$ to be a triple root for $g$, hence a double root for $f$, which is excluded by the hypotheses.