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How many real solutions are there for the equation $$x^{3}-300x=3000$$

Attempt

There is a solution here I found, without using any algebraic factorization: https://youtu.be/y845hT7aYAQ and this seems to be most efficient.

But how about one that uses algebraic factorization, or without using derivative?

One method maybe:

$$(x+a)(x+b)(x+c)=x^{3}-300x-3000$$

Then solve $a,b,c$. LHS is

$$ x^{3} + (a+b+c)x^{2} + (ab+ac+bc)x +abc$$

So we have

$$a+b+c=0$$ $$ab+ac+bc=-300$$ $$abc=-3000$$

Using 3rd equation on the 2nd:

$$ -3000(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})=-300$$

Or $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}=\frac{1}{10}$$

Then $a=-(b+c)$ so $$(b+c)bc=3000$$

By symmetry we also have

$$(a+c)ac=3000$$ $$(b+a)ab=3000$$

Redsbefall
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  • Another method is using Sturm's theorem. –  Jun 05 '22 at 15:32
  • There are many more ways to prove that there is only $1$ real solution: are you specifically looking for a factorisation based approach? –  Jun 05 '22 at 15:36
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    You can use the third degree polynomial equation, but I don't think no one remember it, so maybe it is not the most efficient method. Trying to compute the exact roots without the formula seems prety dificult. Observe that all the equations you wrote can be solved in $\mathbb{Z}$ easily, but not in $\mathbb{R}$ (at least in a simple way) – Marcos Jun 05 '22 at 15:38
  • You can simply find the discriminant and show that it is negative. –  Jun 05 '22 at 15:38
  • https://math.stackexchange.com/questions/2157643/how-can-i-solve-the-equation-x3-x-1-0/2157645#2157645 – lab bhattacharjee Jun 05 '22 at 17:29
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    Each cubic with real coefficients has one real root. There is a way to use the mean value theorem to determine if there is another root – Тyma Gaidash Jun 05 '22 at 22:07

2 Answers2

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This takes some insight, but if you change the $3000$ to $2000$, then you need to factor the polynomial

$$x^3-300x-2000 = (x-20)(x+10)^2$$

which you can do by rational root theorem. Since you have a double root at $x=-10$, you know the graph doesn't cross the $x$-axis, but rises to it, touches, and turn around. Therefore, when you subtract $1000$ from the polynomial, you know that the local maximum is well below the $x$-axis.

B. Goddard
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By Descartes rule of signs, we have one positive root. Zero isn’t a root. Now suppose there are negative roots, then using $x \mapsto -x$, these are positive roots of $x^3+3000=300x$.

However by AM-GM,
$x^3+2000+1000> 300\sqrt[3]2\,x>300x$, hence there are no negative roots.

Macavity
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