# Cubic positive solution exists analysis

1. Jun 20, 2017

### binbagsss

1. The problem statement, all variables and given/known data

Question:

For $\Lambda^3+\Lambda^2+A=0$ , show that for $A<0$ there is some real positive $\Lambda$ which solves this

2. Relevant equations

3. The attempt at a solution

Attempt:

Write $-A=\hat{A}$, then $\hat{A} \in (0,\infty)$
and $\Lambda^3+\Lambda^2=\hat{A}=\Lambda^2(\Lambda+1)$

Since $\Lambda \in (-\infty, \infty)$ then $\Lambda^2 \in (0,\infty)$, and so since $\hat{A} \in (0,\infty)$ then $(\Lambda-1) \in (0,\infty) \implies \Lambda \in (-1,\infty)$

So this shift by $-1$ means that $\Lambda$ is not guaranteed to be positive?

Solution attached:

I understand the solution after the 'claim' $\Lambda \in (0,\infty)$.

However I don't understand where this comes from, this is part of what we are trying to show and we only have $\Lambda$ is real?

2. Jun 20, 2017

### Staff: Mentor

Your post would be easier to read if you followed the lead of the material you posted, and used lower-case lambda ($\lambda$) instead of upper-case lambda ($\Lambda$). At first I misread $\Lambda$ as A. In TeX, the lower-case Greek letters have spellings that start with lower-case letters; i.e., \lambda versus \Lambda, and \theta versus \Theta.

Consider $f(\lambda) = \lambda^3 + \lambda^2 + A$, with A < 0.
$f'(\lambda) = 3\lambda^2 + 2\lambda$
On what intervals is f' positive? zero? negative? On these same intervals the original function will be increasing, horizontal, or decreasing, respectively. This should give you a good idea of where any zeroes of the graph of f are.

BTW, two of the intervals in the posted image use nonstandard notation-- $(0, \infty]$. I've always seen these written as $(0, \infty)$ when an interval extends infinitely far in either direction.

3. Jun 20, 2017

### LCKurtz

I don't think it even requires calculus. Write it as $f(x) = x^3+x^2 -a$ where $a>0$. Then $f(0) = -a<0$ and $f(x) > 0$ for $x>a$.

Last edited: Jun 20, 2017
4. Jun 22, 2017

### SqueeSpleen

It doesn't requiere a lot of analysis, but you need to use intermediate value theorem (Bolzano's).

5. Jun 22, 2017

### WWGD

Actually, this is true for any A, since Complex roots come in pairs, i.e., if z is a root, so is its conjugate z^ , so , by parity alone, there will be a Real root no matter the value of A.

6. Jun 22, 2017

### LCKurtz

Yes, but will it be positive?

7. Jun 22, 2017

### WWGD

Positive... you got me for not reading carefully; top-down processing, my bad. Good point. Maybe I should just delete my post?

8. Jun 22, 2017

### LCKurtz

Nah. I wouldn't worry about it. It happens a lot to all of us.

9. Jun 22, 2017

### SqueeSpleen

Also, some of the algebraics proofs of fundamental theorem of algebra uses the intermediate value theorem in one step (at least the ones that I know with Galois Theory). Then there's are complex analysis proofs that also uses analysis. I think that the less analysis one is the one that only uses the fundamental group of the circle, but intermediate value theorem is more a topological theorem than an analysis one. So I think that using fundamental theorem of algebra to avoid using bolzano's theorem is like a snake bitting it's own tail.

10. Jun 23, 2017

### LCKurtz

Depending on the level of course in which the intermediate value theorem is used, it may very well be that the IVT is taken as "obvious" or an unproved given statement. A high school algebra student may "know" a "continuous" graph can't go from positive to negative without hitting the $x$ axis.

11. Jun 23, 2017

### SqueeSpleen

Yes, you simply justify the afirmation talking about the graphic. But I was saying that it makes sense to put analysis in the tittle, due to that theorem. Of course, it's not mandatory.