# Showing that $m^3+4m+2=0$ has only one real root

I actually have to find the number of real roots of $m^3+4m+2=0$ for a conic sections question in which $m$ is the slope of the normal. The answer is that there is only one real value of $m$, and therefore only one normal.

How did they get this? How can I find out how many real roots such a cubic equation has without spending too much time on it?

• $f(m) = m^3 + 4m + 2$ is a strictly increasing function, so it can only cross $0$ one time. – Doug M May 24 '18 at 1:17
• Do you need to show it without calculus? – G Tony Jacobs May 24 '18 at 1:19
• @GTonyJacobs I would prefer to do so – Hema May 24 '18 at 1:26
• The answer by David K below doesn't use calculus. :) – G Tony Jacobs May 24 '18 at 3:36
• Actually, all of the answers except for mine eschew calculus :) – G Tony Jacobs May 24 '18 at 3:37

The derivative of this function, $3m^2+4$, is always positive, so the function is always increasing. An increasing function on the real line cannot have more than one zero.

The function $f_1(m) = m^3$ is never decreasing. The function $f_2(m) = 2$ is constant. The function $f_3(m) = 4m$ is always increasing.

Add them together, you have a function $f(m) = f_1(m) + f_2(m) + f_3(m)$ that is always increasing. Once it passes through zero it cannot return to zero again.

• I think you mean $f_3(m) = 4m$ – Alex Reinking May 24 '18 at 2:12
• @AlexReinking I could also have written $4f_3(m)$ later--but you're right, $4m$ is what I meant. – David K May 24 '18 at 10:32
• I would have said that $x\mapsto x^3$ is an increasing function because $x_1<x_2\,\Rightarrow\,x_1^3<x_2^3$. – Lubin May 25 '18 at 0:55
• @Lubin I agree. I wrote "never decreasing" (which is implied by "increasing") in order to avoid any possible question about what happens at zero. (Answer: $x<y \implies x^3<y^3$ there too, of course!) I may have been overthinking it. – David K May 25 '18 at 1:21

It's plain that there will be no positive roots. And $f(-m)=-m^3-4m+2 \quad$ leads to one sign change, therefore 1 negative root.

You don't need derivatives to show this. As $m$ gets bigger, $m^3$ also gets bigger, and so does $4m.$ Thus if there is a zero at one value of $m,$ then the value of the function at all larger values of $m$ is more than $0$ and the value of the function at all smaller values of $m$ is less than $0,$ so there cannot be more than one zero.

But there must be at least one zero, since the function approaches $+\infty$ as $m\to+\infty$ and the function approaches $-\infty$ as $m\to-\infty.$

People explained how to do this in this particular situation, but to do this for general cubic polynomials, you need to look at the cubic discriminant, which, for $$ax^{3}+bx^{2}+cx+d$$, is:

$$b^{2}c^{2}-4ac^{3}-4b^{3}d-27a^{2}d^{2}+18abcd$$

For real cubic polynomials, the discriminant is non-negative if and only if the roots are all real.

A common special case (like yours) is $$a = 1$$ and $$b = 0$$, for which this expression reduces to just

$$-4c^3-27d^2$$

In your particular case, $$c = 4$$ and $$d = 2$$ are both positive, so you don't need to evaluate anything—it is obvious that the discriminant is negative, and hence the function has exactly one real root.

### Why does this work?

The general definition of a discriminant is a product of squares, so it can only be negative if the polynomial or the its roots are non-real. The cubic discriminant can be derived from the general definition, and will therefore have the same properties.

Note that $m\ne 0$ and: $$m^3+4m+2=0 \iff m^2+4=-\frac 2m.$$ $f(m)=m^2+4$ is a parabola opening up with vertex at $(0,4)$ and $f(m)=-\frac 2m$ is a hyperbola with one branch in the second quarter. The two conic sections intersect at one point ($m\approx-0.47$) in the second quarter.

The key to finding whether the roots are real or not is that you have to look at the sign of the coefficients. Here, m^3 and m together has have to have value which is equal to 2. Hence, there can only be one point or a number where the equation achieves the value of zero. What you can really do is solve the equation using root of cubit equation method and then use the calculator to find the solution. Here, only one root is real and other two are imaginary and involves (i).

The solution is -0.47346580772912616.

Since $m^3+4m+2\gt0$ if $m\ge0$, any real roots must be negative. If it has more than one real root, then all three roots are real. But by Vieta's formula, the sum of the roots is $0$, the coefficient of the (missing) quadratic term, which is not possible for the sum of three negative values. So there is only real root.

Here is an approach that doesn't require calculus at all:

$$m^3+4m+2=0$$

Now, substitute $$x=k-\frac{4}{3k}$$

$$\therefore m^3+4m+2=\bigg(k-\frac{4}{3k}\bigg)^3+4·\bigg(k-\frac{4}{3k}\bigg)+2$$

$$=\bigg(k^3-3k^2·\frac{4}{3k}+3k·\frac{16}{9k^2}-\frac{64}{27k^3}\bigg)+4k-\frac{16}{3k}+2$$

$$=\bigg(k^3-4k+\frac{16}{3k}-\frac{64}{27k^3}\bigg)+4k-\frac{16}{3k}+2$$

$$=k^3-\frac{64}{27k^3}+2=0\iff \bigl(k^3\big)^2+2k^3-\frac{64}{27}=0$$

Substituting $$z=k^3$$ $$z^2+2z-\frac{64}{27}=0$$ Thus $$z_{1/2}=-1\pm\sqrt{1+\frac{64}{27}}=-1\pm\sqrt{\frac{91}{27}}$$

Can you continue from here?

Remark

Summing up, what we've done here is to obtain a quadratic equation from the cubic one through clever manipulations. This might not be the best method for this exercise, since it is very easy to note with or without calculus, that the function is strictly increasing...