# Values of c for which quartic curve intersects a line

1. Mar 21, 2017

### issacnewton

1. The problem statement, all variables and given/known data
For what values of c is there a straight line that intersects the curve $y = x^4+c x^3+12x^2-5x+2$ in four distinct points ?

2. Relevant equations
Concept of concavity, Vieta's formulas (link)

3. The attempt at a solution

Suppose a straight line $y = mx+b$ intersects this curve. Then to find the points of intersection, we equate the two. $x^4+c x^3+12x^2-5x+2 = mx+b$. This leads to $x^4+c x^3+12x^2-5x+2 - mx-b = 0$. Now this is a polynomial of degree 4. If the line intersects the curve in 4 distinct points, then above equation has 4 distinct roots. Let $r_1,r_2,r_3,r_4$ be 4 distinct roots of this polynomial. Now we use Vieta's formulas, which give the relationship between the coefficients of a polynomials and its roots. For a polynomial of degree 4, we get $r_1+r_2+r_3+r_4 = -c$ and $r_1r_2 + r_2r_3 + r_3r_4+r_4r_1+ r_2r_4 + r_3r_1 = 12$. Since roots are distinct, we have $\sum\limits_{i>j}~ (r_i - r_j)^2 > 0$. Now expanding this sum leads to $$3\sum_i r_i^2 - 2\sum_{i>j} r_i r_j > 0\cdots\cdots (1)$$ Now squaring of $r_1+r_2+r_3+r_4 = -c$ leads to the following $$\sum_i r_i^2 + 2\sum_{i>j} r_i r_j = c^2 \cdots\cdots (2)$$ Its been given to us that $\sum\limits_{i>j}r_i r_j = 12$. Plugging this in equation $(2)$, we get $$\sum_i r_i^2 = c^2 - 24 \cdots\cdots (3)$$ And plugging this in equation $(1)$, we get $$3(c^2 - 24)-2(12) > 0$$ This leads us to inequality, $c^2 > 32$. So we have $|c| > 4\sqrt{2}$. So this must be true for the line to have 4 intersects with this curve. There is another approach using the concept of concavity from Calculus. I argue that if a line is to intersect this curve at 4 distinct points, then the curve changes the concavity at two distinct points. This means that the curve has two inflection points. This further means that if $y = f(x)$, then $f''(x) = 0$ has two distinct roots. Now $f''(x) = 0$ leads to $2x^2+cx+4 = 0$. Since there are two distinct roots, the discriminant must be positive. This again leads to $c^2 > 32$, which is $|c| > 4\sqrt{2}$.

Do you think the solution is correct ?

Thanks

2. Mar 21, 2017

### Staff: Mentor

I get the same result, using the second approach (which is much easier).

3. Mar 21, 2017

### issacnewton

mfb, I wanted to use the first approach as it does not involve Calculus and so can be included in Pre Calculus class. It just uses algebra. But second approach is shorter.

4. Mar 22, 2017

### haruspex

Each method only obtains a necessary condition on c. Neither shows sufficiency.

5. Mar 22, 2017

### Staff: Mentor

I don't see how you can avoid the existence of such a line if you have two inflection points. A line crossing the curve twice between the two inflection points will also cross it again outside (once for each side).

f''(x)=0 leads to inflection points only if f'''(x) $\neq$ 0 (for 4th order polynomials), but the existence of two separate zeros of a parabola guarantees this condition.

6. Mar 22, 2017

### issacnewton

haruspex, By necessary condition you mean the antecedent ? So if $|c| > 4\sqrt{2}$ then we are sure that there is a line which will intersect at 4 distinct points. But if there is a line which intersects the curve at 4 distinct points, then we might have $|c| \leqslant 4\sqrt{2}$. Is that what you are saying ?

7. Mar 22, 2017

### Staff: Mentor

Here is a third approach. Rewrite y:

$$y=\left(x+\frac{c}{4}\right)^4 + \left(12-\frac{6}{16}c^2\right)\left(x+\frac{c}{4}\right)^2 + d\left(x+\frac{c}{4}\right) +e$$
We don't care about d and e here: There is a line with 4 intersections if and only if there is one for d=e=0, simply be adjusting the line.
We also don't care about shifting x, because we can shift our line accordingly.
There is a line with 4 intersections if and only if
$$y=x^4 + (12-\frac{6}{16}c^2)x^2$$
has such a line. That is happening if and only if c^2 > 32, the same condition as before.

8. Mar 22, 2017

### issacnewton

mfb, I don't understand your reasoning here. How did you come up with that form of y ?

9. Mar 22, 2017

### Staff: Mentor

If you expand all the terms, you'll see that the result is the same. It is a standard method to get rid of the second-highest order, shown in more detail here.

10. Mar 22, 2017

### haruspex

Sure, but the answer is incomplete unless it is shown that c2>32 implies two inflection points and that this in turn implies four distinct intersections.

11. Mar 22, 2017

### Staff: Mentor

You can add a small sentence to make it explicit that f'''(x) is non-zero at the two points, sure.

12. Mar 23, 2017

### epenguin

But, for the first part, that is a positive discriminant (of f''(x) ) which does imply two such points.

For the second part, informally, here the coefficient of x4 is positive. The quartic has to have a minimum somewhere. At that point f' = 0, f'' is positive.

Now take only the case where there is only one minimum, I think the other one is easier.

We can draw a line tangent to the inflection point nearest the minimum, and this tangent must intercept the quartic curve at another point. At this point f'' is changing from positive to negative as we get further from the minimum. We can just move the tangent down a little bit so that it will intersect the quartic in three points. After which (I guess I'm not required to be complete in homework help ) it is easy to see that it must intersect in four points.

Last edited: Mar 23, 2017
13. Mar 25, 2017

### issacnewton

Hello mfb, I was busy for 2-3 days. Now I have time, I was thinking about your post #7. Using the link given by you, I was able to understand how you removed the cubic term. But now you say that "There is a line with 4 intersections if and only if there is one for d=e=0, simply be adjusting the line." Why do you need to make d and e equal to zero. We still have a quartic equation, the roots of which will depend upon all the coefficients. Can you explain ? I never worked with quartic equations in algebra before.

14. Mar 26, 2017

### Staff: Mentor

I simplified the equation enough to directly see if a line with 4 intersections exists. That is much easier to see with the linear and constant term gone. As discussed, the existence of these terms doesn't matter. If there is a line with 4 intersections without these terms, then there is also one with these terms present.

15. Mar 26, 2017

### issacnewton

I am not sure I am following the argument here. Can we also delete the quadratic term here ?

16. Mar 26, 2017

### Staff: Mentor

No.
You can ignore the linear and constant term because the straight line will have those terms. Instead of having a curve "+5x", you can subtract 5x from the straight line expression to get the same intersection points. You cannot do that with the quadratic term as the straight line cannot get a quadratic term.

A side-product of this consideration: For every quartic polynomial function, there is always a parabola intersecting it at 4 points.