Tuesday, October 10, 2006

Find the Roots

Here is a fun problem. It was in the Pre-Calculus book that the local highschool uses, so my dad teaches from the book and I tutor from it. It was in a calculator-use section that asked for an estimation of roots only, not an algebraic solution, but just for fun, why not solve it algebraically? :-D I admit that Father Dear discovered the key to solving it, though I carried it through and found the ghastly solutions :-P.

Find all roots of the equation x^3 - 3*x^2 + 3*x + 2 = 0.

You must use high school mathematics concepts only. In other words, don't use the cubic formula!!!


John Dekker said...

What about using high school mathematics concepts to derive the cubic formula?

Susan said...

Nope. Pretend you've never heard of the cubic formula.

Ashley said...

Okay, so I haven't taken math since high school, which means I should be able to use all of my knowledge right? And since I know what the cubic formula is, can I use it? :-)

I decided to be ambitious because I always did like solving for x, but now I can't remember if I'm doing this right, and I have a headache so I gave up. My math teacher would be sorely disappointed in me. (She always wanted me to major in math in college!)

Adrian C. Keister said...

The coefficients look remarkably like those required for a perfect cube, except the constant term is off.

So: write the constant term in a funny way, as follows.

x^3 - 3x^2 + 3x + 2
=x^3 - 3x^2 + 3x - 1 + 3
= (x-1)^3 + 3
= 0

And now, yank the 3 to the other side:

(x-1)^3 = -3.

x-1 = (-3)^(1/3),
or x = 1 - 3^(1/3).

Next, we factor this fellow out, which should leave us with a quadratic.

The way I factored it was to simply set

x^3 - 3x^2 + 3x + 2
=(x-1+3^(1/3))(x^2 + bx + c).

The quadratic has to be monic, so this seemed efficient. You can multiply the RHS out and equate coefficients of like powers of x. The results (and I checked them, they work) are as follows:

b = -2 - 3^(1/3)
c = 2/(3^(1/3) - 1).

I chose b and c thus so that I wouldn't get confused with the quadratic formula. In the quadratic formula for this problem now, a = 1.

Thus, the quadratic formula now reads:

x = [(2 + 3^(1/3) +- sqrt((2 + 3^(1/3))^2 - 8/(3^(1/3)-1))]/2

= (1 + 3^(1/3)/2 +- 1/2 * sqrt([(4 + 4*3^(1/3) + 3^(2/3))(3^(1/3)-1) - 8]/(3^(1/3)-1))

= 1 + 3^(1/3)/2 +- 1/2 *
sqrt( 3(3^(2/3)-3) / (3^(1/3) - 1) )

= x

These two roots, plus the bolded one above, are the three roots of the cubic. There can be no others, so we have found them all.

Is this kosher? Did I break out and use advanced math that no high school student would know? ;-)]

In Christ.

Susan said...

You used the cubic formula in high school, Ashley??? Or did you just hear of it? I really appreciate that you at least *wanted* to solve the problem. I'm touched :-D.

Adrian, your method is correct. *applause* Father Dear also came up with the perfect cube pattern trick. And you worked it all down correctly, although it's easier to synthetically divide using your first root. Not only does this give you a compressed polynomial that is rationalized *ahem*, but it's also easier to work through the quadratic formula afterwards. It's still very nice work, I must say.

But it's really bothering me that your answer is not simplified, or even rationalized. If you did that, it would uncover a rather important feature of the second and third roots (or graph it to find the same feature). I refuse to give full credit for your answer as it currently stands ;). So I would say your method is fine, but your answer is not kosher. By the way, that's really funny that you use that term, because Mother Dear and I describe methods as "kosher" and "non-kosher" also, but I've rarely heard others use it.

Ashley said...

I actually did start solving the problem! I didn't get very far before I started to doubt my own math.

Actually, I'm not sure we used the cubic formula in high school. The name sounded familiar, but now I'm all confused. Ah! Too much math!! You should post simpler questions next time. Like 3x+2=8. I know that one. :-)

John Dekker said...

I must admit, Ashley, I was quite surprised - the cubic formula is very complicated. And no, I couldn't remember how to derive it.

Adrian, well done. It was very warm last night and I couldn't sleep, so I got up and did some mathematics :) But I didn't see it! I tried so many things, but not the difference of perfect cubes. It's been far too long...

Susan, thanks for posting this on your blog. It's good to get back into maths. (Yes - that's what we call it - not "math")

By the way, once we have the difference of cubes, we can also factorise by observing that a^3 - b^3 = (a - b)(a^2 + ab + b^2).

I guess what Susan meant by not getting full credit was that two of the roots are complex.

[I must admit that I did cheat at one point and use calculus, and that told me that we had a point of inflection at x=1. This means that we have only one real root...]

If we look at the bits that Adrian put into bold, we can see that since 3^(1/3) - 1 is postive and 3^(2/3) - 3 is negative, the bit inside the square root is negative, which does not produce a real solution.

But I vote we give Adrian the benefit of the doubt and assume that he knew that. :)

Looking forward to the next problem...

Adrian C. Keister said...

Ja, I didn't bother to check to see if the two last roots I got were complex-conjugate pairs. How would you rationalize my answers? (Please don't interpret me saying that in a defensive tone, by the way!) You can't get rid of the cube root in the denominator, can you?

Actually, John, the existence of an inflection point does not guarantee only one real root. Counterexample:

(x-1) x (x+1) = x^3 - x.

The first derivative is 3x^2 - 1, and the second derivative is 6x. The second derivative is zero at the origin, and is negative for negative x and positive for positive x. Thus the concavity changes at the origin, and we have an inflection point. However, you can just check that -1, 0, and 1 are the roots of this cubic.

What would guarantee only one real root? If the first derivative is always positive or always negative, that would be sufficient (but not necessary) conditions for only one real root. In our case, 3x^2 - 6x + 3 is the first derivative. You can see that this is 3(x^2 - 2x + 1), which is always positive. Hence, there is exactly one real root. (It's a odd-order polynomial, so it's guaranteed, with real coefficients, to have at least one real root).


In Christ.

Susan said...

Problem #2: 3x + 1 = 7. Find all solutions.


Interesting about "maths," John. That actually makes more sense, since it's an abbreviation of mathematics. . . Aussies re still weird ;). And, fine, I'll give Adrian the benefit of the doubt. . . at least mostly. I am think a 90%.

Haha, Adrian. I actually did read your question about rationalizing in a defensive tone at first, albeit mocking :). But then I immediately saw what you mean. I admit I did not previously know how to rationalize binomial cube roots, but I looked it up yesterday in order to check your answers against mine :-D. The key is to use the difference of cubes pattern. For a binomial with a cube root, just complete the pattern with a trinomial that will give you a difference of cubes. Nifty, eh? But that trick isn't needed if you just sythetically divide to begin with. . .

Ashley said...

Ooh! Oooh! I know, x=2

Sweet, I feel a sense of satisfaction in my life. Who knew that math could bring this? *light bulb goes on in head*

Susan said...

That was awesome, Ashley. I'm so proud. You deserve a picnic supper at Stone Mountain for your accomplishment. What say you? ;)

Adrian C. Keister said...

Reply to Susan.

You are hard to please, aren't you? You should have been named Adrienne. ;-)] I'm going to be just as hard on you if you attempt the problems on my calculus post. Well, I understand the procedure required to rationalize the roots, so I'm not going to do it. So there! *grins* That's the sort of thing I'd just plug the numbers into a calculator in order to find out what sort of thing they are (real, complex conjugate pairs, etc.) Hey, check out my calculus post. Enjoy!

In Christ.

Susan said...

Haha. No, I'm not an Adrienne, I'm a Susan. But yes, I'm hard to please. Although I hardly think requiring you to put your answer in simplified, complex form is being difficult. You couldn't even tell from your answer that the 2nd and 3rd roots were complex! That's a major issue. You're refusing to finish your work? Difficult boy. I've docked your grade to a 69.

Radagast said...

Nice post!

In response to Adrian, if there's only one real root r, then the polynomial is (x-r)(x^2+bx+c) and by the quadratic formula you're always going to get complex-conjugate pairs.

Interesting that solving quadratic equations is so easy (the Babylonians could solve the non-complex cases), but cubics are so hard (the formula was discovered in the 16th century).

Susan said...

Radagast, certainly if there's only one real root r then a cubic is always going to get complex-conjugate pairs. I believe what Adrian and John were discussing was how one could algebraically discover if there was only one real root, not if that nugget was helpful once found :).

Interesting you brought that up, though, because that is actually what generated my interest in this problem. I meant to recount this before. As I mentioned in the original post, this was a problem I pulled from a high school Pre-Calculus book, and it was under a calculator-estimation section, not an algebraic solution section. . . But I didn't see that at first. My tutoring student was picking random problems from the chapter review to study for a quiz, and he chose thise one.

He first checked all the possible rational zeros (+-1, +-2) by synthetic division, and none of them worked. Usually cubics aren't hard to solve: just find a rational zero and use the quadratic equation to finish it. But I was genuinely perplexed that there were no rational zeros. I was thinking, "There can't be 3 imaginary solutions! - I was visually graphically why this must be the case." I even graphed it on a calculator, but could only tell it had a negative real solution.

Then I looked at the last coefficient (2) and wondered how the real solution could be irrational. Complex solutions always come in pairs, but also irrational solutions usually do. But obviously not here, which created the problem of rationalizing the polynomial. Then I realized that the complex roots must also be irrational, and necessarily rationalize the polynomial when multiplied by the irrational real root. Bleh, what an ugly problem. But a very interesting method of solving, which is beautiful.

Adrian C. Keister said...

Reply to Radagast.

Well, given only one real root of a cubic, you will complex conjugate pairs since there are real coefficients. If you had complex coefficients, all bets are off. In a lot of ways, I prefer the complex numbers; proofs are easier. Complex analysis is a very beautiful theory. Powerful theorems, easy proofs. Most of that is because there is only one infinity, unlike the reals with its two infinities.

Reply to Susan.

Mathematicians are inherently lazy, don't you know? You've probably heard the one about the engineer sleeping. A fire alarm woke him up, at which point he got up, rushed around, found a fire extinguisher, and put out the fire. He went back the sleep. A little while later the fire alarm went off again and woke up the mathematician. He sat up in bed, thought for a moment, said to himself, "There is a solution," and went back to sleep.

Or there's the one about the company that wanted to hire a mathematician and not and engineer. So they devised this test. The test consisted of two rooms equipped with the following: stove, kettle, and table. In the first room, the kettle is on the table. So the engineer walks in, moves the kettle from the table to the stove, and boils the water. The company resets the room, and in walks the mathematician. He moves the kettle from the table to the stove and boils the water. Then they come to the second room. This time the kettle is on the floor. The engineer walks in, moves the kettle from the floor to the stove and boils the water. They reset the room, and in walks the mathematician. He moves the kettle from the floor to the table and says, "We've reduced this to a problem already solved."

And so I have to confess that although I'm ok at algebra, it's not my favorite math by a long shot. I try to avoid doing as much of it as possible. Calculus is my thing.

In Christ.

John Dekker said...

We've reduced this to a problem already solved


Radagast said...

I'm with John... I'd work out how many real roots a cubic has with calculus -- either finding a point of inflection (as in this case) or finding maxima and minima.

What do you mean, kids don't do calculus in high school? They SHOULD. ;)

Susan said...

Those math jokes were hilarious, Adrian :). I remember finding a treasure trove of such jokes in the thick of my semester taking Abstract and Non-Euclidean Geometry, and they were considerably amusing - and true. My favorite is the "How to Prove It" guide for lectures:


How to prove it. Guide for lecturers.

Proof by vigorous handwaving:
Works well in a classroom or seminar setting.

Proof by forward reference:
Reference is usually to a forthcoming paper of the author,
which is often not as forthcoming as at first.

Proof by example:
The author gives only the case n = 2 and suggests that it
contains most of the ideas of the general proof.

Proof by omission:
"The reader may easily supply the details"
"The other 253 cases are analogous"

Proof by deferral:
"We'll prove this later in the course".

Proof by picture:
A more convincing form of proof by example. Combines well
with proof by omission.

Proof by intimidation:

Proof by seduction:
"Convince yourself that this is true!"

Proof by cumbersome notation:
Best done with access to at least four alphabets and special

Proof by exhaustion:
An issue or two of a journal devoted to your proof is useful.

Proof by obfuscation:
A long plotless sequence of true and/or meaningless
syntactically related statements.

Proof by wishful citation:
The author cites the negation, converse, or generalization of
a theorem from the literature to support his claims.

Proof by reduction to the wrong problem:
"To see that infinite-dimensional colored cycle stripping is
decidable, we reduce it to the halting problem."

Proof by reference to inaccessible literature:
The author cites a simple corollary of a theorem to be found
in a privately circulated memoir of the Slovenian
Philological Society, 1883.

Proof by importance:
A large body of useful consequences all follow from the
proposition in question.

Proof by accumulated evidence:
Long and diligent search has not revealed a counterexample.

Proof by mutual reference:
In reference A, Theorem 5 is said to follow from Theorem 3 in
reference B, which is shown to follow from Corollary 6.2 in
reference C, which is an easy consequence of Theorem 5 in
reference A.

Proof by metaproof:
A method is given to construct the desired proof. The
correctness of the method is proved by any of these

Proof by vehement assertion:
It is useful to have some kind of authority relation to the

Proof by ghost reference:
Nothing even remotely resembling the cited theorem appears in
the reference given.

Proof by semantic shift:
Some of the standard but inconvenient definitions are changed
for the statement of the result.

Proof by appeal to intuition:
Cloud-shaped drawings frequently help here.


I've seen pretty much all of these methods by my college professors :).

Susan said...

Then, boys, there's always the plug-it-into-the-graphing-calculator method of finding the number of real roots :-D. Or is that cheating? :)

John Dekker said...

Interestingly enough, I find a lot of these methods of proof coming up in theology. Especially :

cumbersome notation - anything with Ugaritic, Akkadian or Syriac.

wishful citation - almost any reference to Calvin :(

reference to inaccessible literature - David Jobling, “The Canon of the Hebrew Bible As a Literary Work.” Unpublished
paper delivered at the Canadian Society of Biblical Studies Annual
Meeting, University of Windsor, Ontario, 1988.

accumulated evidence - "To date, no one has refuted John Owen's defence of limited atonement."

Adrian C. Keister said...

Now, now, John. Calvin's writings were inspired. I thought everyone knew that! Well, at least all Reformed folk. ;-)]

Ja, I have seen those proof techniques before, both in print and in the classroom. Here's a couple more:

Proof by instant erasure.
Helps to write with your chaulk in one hand and erase with the other as you go.

Proof by delegation.
"An exercise for the reader."

Your proof by intimidation has given rise to stories such as the following: a mathematics professor wrote a theorem or fact on the board and commented, "Of course, this is inherently obvious." At which point he stopped and stared at the board for fully ten minutes without moving a muscle. At the end, he said, "Yep, it's obvious."

Have I ever told you the Norbert Wiener stories? Those are the best. Stereotypical absent-minded professor type.

Oh, and using the calculater isn't cheating. As the engineers would say, we would get results close enough for all practical purposes.

In Christ.

John Dekker said...

Well, it's Calvin's inspired status which makes him such a target for wishful citation. So often he's appealed to by people on both sides of a debate.

Susan said...

You mean, Calvin's writings aren't inspired???? I'm feeling like a poor, confused reformed girl right now. *fans self with math book* . . . That is a funny tie between math proofs and theological treaties, John.

Proof by Instant Erasure. Ah yes, Adrian, I've seen that :). And leaving the exercise to the reader. . . that was prevalent in many of my math texts :). Hehe.

I think you may have posted some Wiener anecdotes quite a while ago(?). Yes, here we go. I remember thinking they were funny :). I think I'll take being less smart than him and still functioning as a semi-normal human being ;).

John Dekker said...

Susan and Adrian, you will both love this blog post.

John Dekker said...

Oh, I enjoyed the Wiener stories. I also like proofs that 1 = -1. Here's another one:

if and only if = iff
Cancel i,f,f on both sides:
and only i = 1
Square both sides:
-1 = 1.

Susan said...

Those were funny, John. The limit especially tickled me :). We had a good laugh over that at breakfast. And the problem that said "find x" reminded me of something one of my math education professors loved to say:

Most students spend their whole high school math career searching for "x". . . and they never find it!

une_fille_d'Ève said...

This is Susan's father.

Actually the limit of 1/(x-8) as x approaches 8 is undefined, not infinity.

However, the limit of 1/(x-8)^2 as x approaches 8 would be infinity.

Adrian C. Keister said...

Reply to Mr. G.

So as to restore mathematical accuracy, I have seen that funny as a one-sided limit, from the right. Then you get the right thing. :-)]

In Christ.

Ashley said...

There are too many comments on this post.

Susan said...

Oh no! Did we exceed the Blogger comments-per-post limit, Ashley???? What is the limit, so we don't make the same mistake again :)???

John Dekker said...

It doesn't help if people make comments just for the sake of it. ;)