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Suppose we have $x^3 + \alpha x + \beta = 0$

Problem:

the cubic equation has one real root if $\alpha > 0$ and three real roots if $4 \alpha^3 + 27 \beta^2 < 0 $.

Try:

Let $f(x) = x^3 + \alpha x + \beta$. One has that $f'(x) = 3 x^2 + \alpha $. We have critical points when

$$ 3x^2 + \alpha = 0 \iff x^2 = - \frac{ \alpha }{3} $$

Clearly, if $\alpha > 0$, then we dont have critical points and $f'(x) > 0$ thus always increasing. Since $\lim_{x \to \infty} f(x) = \infty$ and $\lim_{x \to -\infty} = - \infty$, then there must be some $c$ such that $f(c) = 0$, so we have one real root.

Now, if $\alpha < 0$, then we have critical points $x = \pm \sqrt{ -\alpha/3 } $. but, here I am stuck. Any help would be greatly appreaciated.

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  • $\begingroup$ are you assuming that $\alpha$ and $\beta$ are real numbers? $\endgroup$
    – Mirko
    Commented Dec 6, 2016 at 3:04
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    $\begingroup$ Just show that nonreal roots come in conjugate pairs, i.e. if $z = a + bi$ is a root of $x^3 + \alpha x + \beta$, then so is $\bar{z} = a - bi$. $\endgroup$ Commented Dec 6, 2016 at 3:09
  • $\begingroup$ Have you tried decomposing into a quadratic, times a linear function and looking at the discriminant of the quadratic? $\endgroup$
    – wesssg
    Commented Dec 6, 2016 at 3:47

3 Answers 3

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$x=-\sqrt{-\alpha\over3}$ is the maximum, $x=\sqrt{-\alpha\over3}$ is the minimum.

If we want three real roots it must be:

$$[(-t)^3-\alpha t+\beta][t^3+\alpha t+\beta]<0,$$

(where $t=\sqrt{-\alpha\over3}$). In fact we know that if $f(x)=x^3+\alpha x+\beta:$ $$\lim_{x\to\pm\infty}f(x)=\pm\infty,$$ $f'(x)>0$ if $x<-\sqrt{-\alpha\over3}$; $f'(x)<0$ if $-\sqrt{-\alpha\over3}<x<\sqrt{{-\alpha\over3}}$ and $f'(x)>0$ if $x>\sqrt{-\alpha\over3}$, so if the above inequality is true, then the maximum must be positive and the minimum negative and the function intersects the $x-axis$ at three points (formally you can apply the Bolzano theorem), i.e. we must have three real roots.

If you calculate the expression you get:

$$-t^6-\alpha t^4-\beta t^3-\alpha t^4-\alpha^2 t^2-\alpha\beta t+\beta t^3+\alpha \beta t+\beta^2<0$$ $$-t^6-2\alpha t^4-\alpha^2 t^2+\beta^2<0$$ $$-{\alpha^3\over27}-{2\over9}\alpha^3+{\alpha^3\over3}+\beta^2<0$$ $${4\over27}\alpha^3+\beta^2<0$$

as required.

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  • $\begingroup$ Why must $((-t)^3 - \alpha + \beta) (t^3 + \alpha t + \beta ) < 0$ if we want three real roots? $\endgroup$
    – user139708
    Commented Dec 6, 2016 at 5:21
  • $\begingroup$ @Jibarito I added something...I hope it is helpful $\endgroup$
    – MattG88
    Commented Dec 6, 2016 at 13:48
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The fundamental thereom of algebra describes the roots of polynomials. In this case , the possibilities for a cubic equation is 1 complex pair (and one real root) or (three real roots). This is a little hand wavy, but you could use it more formally

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Consider Vieta's formula for cubic polynomials:

given $P(x) = ax^3+bx^2+cx+d$ and roots $x_1,x_2,x_3$ for $P(x) = 0$, we know

$$x_1+x_2+x_3 = -\frac{b}{a}$$ $$x_1x_2+x_1x_3+x_2x_3=\frac{c}{a}$$ $$x_1x_2x_3 = -\frac{d}{a}$$

Your polynomial is the special case with $a=1,b=0$

Now assume a solution exists, where two roots are real and the third is imaginary, w.l.o.g. $x_1$ and $x_2$ are real, but $x_3$ is not. Let's say $x_3 = x_{3r} + i \cdot x_{3i}$, with $x_1,x_2,x_{3r},x_{3i} \in \mathbb{R}, x_{3i} \neq 0$.

Now: $$x_1 + x_2 + x_3 = -b / a = 0$$ $$\Leftrightarrow x_1 + x_2 + x_{3r} + i \cdot x_{3i} = 0 + 0 \cdot i$$ Just looking at the imaginary parts, we get $$\Rightarrow x_{3i} = 0$$

Which is a contradiction, thus two real roots is impossible. Since you can find examples with $0,1$ and $3$ real roots, you're done.

And $\alpha, \beta$ can be complex numbers for this.

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