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.
Best Answer
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.