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We know that a quadratic equation has at most two real roots. Now, how many real roots can a cubic equation $x^3 + bx^2 + cx + d = 0$ have? Explain your answer.

I know by the back of my head that a cubic equation has either one real root or three real roots. However, how do I go about proving it? If it's possible, I would appreciate examples to showcase this.

What i have attempted so far:

Since $x^3 + bx^2 + cx + d = 0$ is a cubic polynomial equaation, it is continuous on $[a,b]$, where $a<b$, and differentiable on $(a,b)$. Thus, by Roelle's Theorem, there exists $d ϵ (a,b)$ such that $$ f'(d) = 0 $$ $$ 3x^2 + 2bx + c = 0 $$ Hence, this shows that there exists at least one real root on this cubic equation.

How do I then show it has three real roots as well?

Thanks.

Rodrigo de Azevedo
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idolo
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    $x^2(x-1)$. How many real roots? – imranfat Sep 26 '17 at 17:11
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    The fundamental theorem of algebra says that any polynomial has a unique factorization $a(x-r_1)(x-r_2)(x-r_3)$ Now, those roots could be complex, or real, or roots of multiplicity. But the maximal number of real roots equals the degree of the polynomial. If it has complex roots and real coefficients, complex roots will be found in conjugate pairs. – Doug M Sep 26 '17 at 17:13
  • Well, you can show by example: that it can have 1 ($x(x^2 + 1)$, 2($x^2(x+1)$) or 3($x(x+1)(x-1)$). And you have shown it can't have 0. Can you accept it can't have more than 3? – fleablood Sep 26 '17 at 17:46
  • A cubic equation has at least one real root this can be shown using the Intermediate Value Theorem – Tom Carter Sep 26 '17 at 19:54
  • https://en.wikipedia.org/wiki/Budan%27s_theorem – IV_ Aug 13 '24 at 20:07

3 Answers3

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Let $p(x) = x^3 + bx^2 + cx + d$.

$p(x)$ is continuous, and we know there is a sufficiently large $x^*$ with $p(-x^*) < 0 < p(x^*)$. By the intermediate value theorem, there is at least one point in $(-x^*,x^*)$ where $p(x) = 0$. So $p(x)$ has at least one real root.

If $p(x)$ has four real roots, then the following must be satisfied $$ \left[\begin{matrix}x_1^3 & x_1^2 & x_1 & 1 \\x_2^3 & x_2^2 & x_2 & 1 \\x_3^3 & x_3^2 & x_3 & 1 \\x_4^3 & x_4^2 & x_4 & 1 \end{matrix}\right]\left[\begin{matrix}1 \\ b\\c\\d\end{matrix}\right] = \mathbf{0} $$ Since the determinant of the matrix is $(x_1 - x_2)(x_1-x_3)(x_1-x_4)(x_2-x_3)(x_2-x_4)(x_3-x_4) \ne 0$, this is impossible. $p(x)$ cannot have four (or more) real roots.

So $p(x)$ must have at least one real root, and it can't have four or more. Can it have two or three? Well, $p(x) = x^3 -x^2$ has two real roots, and $p(x) = x^3 - x$ has three real roots, so these are clearly possible. Thus, a cubic can have one, two, or three real roots.

eyeballfrog
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Look at $|x| \to \infty$ for the cubic equation and use the intermediate value theorem to conclude that there is at least on a real solution.

Using this we can rewrite the polynomial $p(x)=(x-x_1)(\alpha x^2+\beta x+\gamma)$

Now as we have a quadratic equation with real coefficients $\alpha x^2+\beta x+\gamma$, we know there are three possible cases.

Case 1: no real solution (in total 1 real root for the cubic).

Case 2: we have two distinct real solutions (in total 3 real roots).

Case 3: We have a real solution with a multiplicity of 2 (in total 3 real roots).

MrYouMath
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Although it is well known the analytical solution to this equation, we present an alternative and closed-form for the solution where it is possible to detail the nature of the roots depending on the coefficients. To do this, I do suggest you take a look at Solutions 1 and 2 which are based on the Lambert-Tsallis function ($W_r(z)$). The first one is related to the general idea to describe the following answer meanwhile the second solution will present a graphical behaviour of $W_r(z)$ for $r$ even.

As you may know the equation $x^3 +bx^2 +cx +d = 0$ can be written as $$y^3 + \big(c-\frac{b^2}{3} \big)y + \frac{2b^3}{27} -\frac{bc}{3}+d=0 $$ with $x=y-\frac{b}{3}$.

We will rewrite the main equation as $$y^3 + a_m\cdot y + a_0=0 $$ with $a_m= c-\frac{b^2}{3}$ and $a_0= + \frac{2b^3}{27} -\frac{bc}{3}+d$.

The previous equation can be written as $$y=\frac{-a_0/a_m}{1 + \frac{W_2(z)}{2}} \implies x= \frac{-a_0/a_m}{1 + \frac{W_2(z)}{2}} -b/3$$ with $z=\big[2\big(\frac{1}{a_m}\big)\big(-\frac{a_0}{a_m}\big)^2 \big]$. Once that $x$ is written in terms of basic operation on $W_2(z)$, one can infer there are 3 different ranges which define the number of real roots as you requested:

  1. $z>0$ which is $a_m>0$ or $3c>b^2$

    On this situation there is only one real root and a pair of conjugated-complex.

  2. $z_b < z <0$ where $z_b = -8/27$.

    At this condition it is embedded $3c<b^2$ and the general equation is $$-2 < \frac{27d+2b^3-9bc}{b^2-3c}\cdot \bigg( \frac{1}{b^2-3c}\bigg)^{1/2}<2.$$

In this case there are three real roots.

  1. $z < z_b$.

    At this condition it is embedded $3c<b^2$ and the general equation is $$\frac{27d+2b^3-9bc}{b^2-3c}\cdot \bigg( \frac{1}{b^2-3c}\bigg)^{1/2}>2$$

On this case, there is also 1 real solution and a pair of conjugated-complex.

ZKZ
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