# When does quartic have only one real solution?

1. Jul 6, 2010

### Gerenuk

Does someone know what condition for the coefficients of the quartic polynomial has to be fullfilled so that it has exactly one real solution (i.e. one real zero)?
I tried looking up the general solution, but I get lost in all these transformations. Is there an easy way to tell?

Last edited: Jul 7, 2010
2. Jul 6, 2010

### Staff: Mentor

If the coefficient of the 4th degree term is positive, the graph of a quartic polynomial comes down from the 2nd quadrant and goes up into the 1st quadrant, very roughly like a U. If this coefficient is negative, the graph comes up from the 3rd quadrant and goes down into the 4th quadrant, something like an upside-down U.

Such a graph can't cross the x-axis an odd number of times (i.e., once or three times). What must happen for the graph to have exactly one real zero?

3. Jul 6, 2010

### Gerenuk

Hmm, my first guess is that the odds terms must be non-zero under some conditions and it also should just touch the x-axis. But how can I translate that into a condition for the coefficients?

4. Jul 6, 2010

### Redbelly98

Staff Emeritus
I don't know the actual answer, but if it "just touches" the x-axis, then that says something about both the function and it's derivative at that point. That suggests two relations as a necessary (though perhaps not sufficient) condition. This may or may not lead somewhere useful here, but you might try looking at that.

5. Jul 6, 2010

### Dickfore

Are the coefficients of the equation real or complex numbers?

6. Jul 6, 2010

### hamster143

Do you want exactly one solution, period, or exactly one solution that is zero?

If you want the only solution to be zero, then, obviously, $a_0$ should be zero, therefore you can rewrite it as

$$f(x) = x(a_1 + a_2 x + a_3 x^2 + a_4 x^3)$$

and the quantity inside the brackets is a cubic that must have at least one root, and it must be zero also, therefore, $a_1$ is zero too:

$$f(x) = x^2 (a_2 + a_3 x + a_4 x^2)$$

and the remaining terms must satisfy either $a_2=0, a_3=0$ or $a_3^2-4 a_2 a_4 < 0$.

In the general case, you can rewrite the polynomial as f(y-c), where c is the only zero, and open the brackets ... it'll be messy ...

7. Jul 7, 2010

### Gerenuk

The coefficients are only real numbers.
I want that the quartic has exactly one real solution. I don't know any solution and want to avoid the complicating procedure for quartics.

8. Jul 7, 2010

### Char. Limit

Here's an example...

$$f\left(x\right) = \left(x+a\right)^2 \left(x^2 + b^2\right)$$

Of the two terms, the first is squared and this is positive for all x =/= -a, and 0 for x=-a. The second term is always positive.

9. Jul 7, 2010

### Gerenuk

OK, that's an example.
But I'm more interested in determining that for a general quartic.

10. Jul 7, 2010

### Dickfore

Then, if z is a root, so is its complex conjugate z*. But, every polynomial of degree n has exactly n complex roots, some of which might repeat. As a consequence, a polynomial with real coefficients of odd power has at least one real root.

If z = 0 is a root, then, the quartic polynomial can be written as:

$$P_{4}(z) = z \, Q_{3}(z)$$

But, $Q_{3}(z)$ is of odd power and with real coefficients, so it must have at least one real root. Since, the only possible real root is 0, it must be that:

$$Q_{3}(z) = z \, R_{2}(z)$$

with $$R_{2}(z)$$ being a polynomial of degree 2 with either no real roots or 0 being its double root. Without any loss of generality (since we are looking for the roots of the polynomials), we can take the coefficients in front of z4 to be 1. We can then write:

$$P_{4}(z) = z^{2} \, (z^{2} + p \, z + q), \ p^{2} - 4 q < 0$$

or

$$P_{4}(z) = z^{4}$$

11. Jul 7, 2010

### Gerenuk

Why should one zero be equal to zero? I have a general quartic.

12. Jul 7, 2010

### Dickfore

Ohhh, I thought that the (zero) in your opening post was the value of the real root. Regardless. If a quartic polynomial $P_{4}(z)$ has only one real root at 0, then the polynomial with a translated argument $P_{4}(z - \alpha)$ has only one real root at some real number $\alpha$. The above conditions become:

$$P_{4}(z) = (z -\alpha)^{2} \, \left[ (z - \alpha)^{2} + p \, (z - \alpha) + q \right], \ p^{2} - 4 q < 0$$

Multiply out and compare with:

$$P_{4}(z) = z^{4} + a_{1} \, z^{3} + a_{2} \, z^{2} + a_{3} \, z + a_{4}$$

to express $\alpha$, p and q in terms of the coefficients. Then, substiute those expressions in the condition $p^{2} - 4 q < 0$. This will give you one condition among the four coefficients $a_{i}, i = 1, 2, 3, 4$.
In the second case

$$P_{4}(z) = (z - \alpha)^{4}$$

we have the fourth power of a binomial, so we have:
$$a_{1} = - 4 \, \alpha \Rightarrow \alpha = -\frac{a_{1}}{4}$$

$$a_{2} = 6 \, \alpha^{2}$$

$$a_{3} = -4 \, \alpha^{3}$$

$$a_{4} = \alpha^{4}$$

So, in this case we have 3 conditions:

$$a_{2} = \frac{3 \, a^{2}_{1}}{8}, a_{3} = \frac{a^{3}_{1}}{16}, a_{4} = \frac{a^{4}_{1}}{256}$$

13. Jul 7, 2010

### Gerenuk

Oh sorry. My fault with the confusing zero.

Now these three equations seem to be too restrictive?! CharLimits example does not obey it?

14. Jul 7, 2010

### Dickfore

Because his example is of the first kind with

$$\alpha = -a$$

$$p - 2 \alpha = 0 \Rightarrow p = -2 a$$

$$\alpha^{2} - p \alpha + q = b^{2} \Rightarrow q = a^{2} + b^{2}$$

Using these values for p and q, the condition is:

$$p^{2} - 4 q = (-2 a)^{2} - 4(a^{2} + b^{2}) = -4 b^{2} < 0 \Leftrightarrow b \neq 0$$

The first case is more general, but algebraically harder.

15. Jul 7, 2010

### hamster143

It seems to me after about 10 min of algebra that the problem of expressing p and q in terms of a_i is as hard as solving the original quartic (which is what we're trying to avoid here).

Let's solve a half-problem first. Can we take a general quartic and determine, without solving it or any other quartic, whether it has a double root, in other words, has the form

$$P_{4}(z) = (z -\alpha)^{2} \, \left[ z^2 + p \, z + q \right]$$

for some $\alpha, p, q$?

16. Jul 7, 2010

### Gerenuk

17. Jul 7, 2010

### trambolin

It should help if you look at the polynomial as a quadratic expression as such
$$P(x) = \begin{pmatrix} 1\\x\\x^2\\ \vdots\end{pmatrix}^T \begin{pmatrix}a_0 &\frac{a_1}{2} &\cdots\\ \frac{a_1}{2} &\ddots &\cdots\\ \vdots & &\ddots\end{pmatrix} \begin{pmatrix} 1\\x\\x^2\\ \vdots\end{pmatrix}$$

So you might say something clever about having one of the eigenvalue zero and the rest being the same sign i.e. the matrix being positive semidefinite or negative semidefinite. So rank drops only by 1. Did not put too much thinking though, but seems doable.

18. Jul 7, 2010

### Staff: Mentor

Knowing that there is just one real root enables you to write the equation as
(x - r)2(ax2 + bx + c) = 0.

Here r is the root that is known. Because the graph of y = (x - r)2(ax2 + bx + c) can only touch the x-axis and not cross it (a quartic that crosses the x-axis has to have a second root where it crosses again), the values of (x - r)2(ax2 + bx + c) must be nonnegative (if a > 0) or nonpositive (if a < 0), and can be zero only when x = r. That should tell you something about the ax2 + bx + c factor.

19. Jul 7, 2010

### Gerenuk

That's certainly true, but it seems I cannot find out a,b,c without knowing the root?!

20. Jul 7, 2010

### hamster143

Here's a new idea.

If your polynomial has form $P(x) = (x - r)^2 (x^2 + px + q)$, then it's true that

$$P(r) = 0$$
$$P'(r) = 0$$

By expanding these and doing some manipulations, you can demonstrate that r would be the root of quadratic equation

$$(8a_2 - 3a_3^3)r^2 + (6a_1-2a_2a_3^2) r + (4a_0 a_3 - a_1 a_3^2) = 0$$

* Solve the equation for r
* Plug the solution(s) back into P to see if they work. If not, exit
* Observe that $p = a_3 + 2r$ and $q = a_0/r^2$.
* Check that $p^2-4q<0$.