Rational Roots of a Quadratic Equation

Here is an interesting excursion to try; take a quadratic equation that you know has rational roots, and then permute the values of A, B, and C through all possible variations. What is the probability that they would ALL have rational roots?

For example, you might try 2x²+7x + 3, which has roots of -1/2 and -3. If you switch to 2x²+3x + 7 there are no real roots at all. In fact, four of the six permutations have imaginary roots, and only 3x²+7x +2 also has rational roots.

If instead, you had used x²+3x - 10 (roots at 2 and -5) you would find three other permutations had rational roots, 3x²+x -10, -10x²+3x + 1, and -10x²+1x + 3 . The other two permutations have real irrational roots.

What happened with the choices you made for A, B, and C (you DID do it didn't you???).
Several questions pop up...

Is it possible for All six of the permutations to have rational roots?

Both my examples have an even number of the permutations that have rational roots. Is it possible that the number of permutations with rational roots could be odd?

Is it possible that a choice of A, B, and C would have two permutations with rational roots, two with irrational roots, and two with imaginary roots? (sort of the ultimate root trifecta)


The answer to the first question is yes, in fact, I can prove (and you probably can too) that if we select A, B and C so that A+B+C = 0, then all six permutations will have rational roots. The roots will be rational if b^2-4*A*C is a square of a rational, so we can substitute -(A+B) for C and see that (B,²-4A(-A+B)=B²-4AB+4A²=(B-2A)²... An unresolved problem at the moment, which I will have to play with some, is whether this works both ways. Is it possible that all six permutations have rational roots when A+B+C is not Zero??? (please advise if you get this before I get back to it..)

I like the second question because it can be solved with a simple appeal to symmetry. Since B²-4AC is equal to B^2-4CA, exchanging A and C will always produce the same value for the discriminant (B²-4AC). So the number of permutations which have rational roots will always be even.

For the Third, I resorted to a computer search and the first I found was with the values of 1, -4, and -5. x²-4x-5 has rational roots, x²-5x-4 has irrational roots, and -4x²+x-5 has imaginary roots.

Can You SEE the solution?

It may have been G. H. Hardy who first stated that mathematics is about finding patterns. In his autobiography he writes, "A mathematician, like a painter or a poet, is a maker of patterns. If his patterns are more permanent than theirs, it is because they are made with ideas." It is an idea I often repeat to my students, and to myself when I am trying to present new ideas to them.

We are at that point in the year where we are working with the Fundamental Theorem of Algebra. We have gotten to the point that most of the students can graph an equation, y=f(x) and looking at the graph write the factored form if the roots are all rational. In the vernacular of the kids, the can see the roots. As we began working with complex numbers, one student lamented that they wished you could see the roots as well on a quadratic with non-real solutions. Several others murmured agreement, and it seemed like a teachable moment, so I turned and said, cryptically I hope, "Well, you can, but you have to recognize the pattern." It is hard to believe they thought that there was something simpler than the quadratic formula, but they are more willing to do ten minutes work with a calculator than two minutes with a pencil, and besides, it is pretty, so I was willing to lead them toward it.

I began by setting up some simple problems. graph the equation and find the vertex, then solve by the quadratic formula (never tell them they are actually practicing, we are discovering exciting, not in the book, kind of stuff here).

They do a few:
x² -2x + 10, with a vertex at (1, 9) had solutions at x= 1 +/- 3i

x² +4x + 5, with a vertex at (-2,1) had solutions at x= -2 +/- 1i

x² -6x + 12, with a vertex at (3,3) had solutions at x= 3 +/- isqrt(3)
AHHH, now they had a clue.....

We do a few more, and they seem to be on top of it, but not one has noticed that ALL the problems had a leading coefficient of one... with only a few minutes to go, I gave them
2x² -4x + 20 (no one notices, it seems, that this is twice the first problem we had done) ... they sketch the graph, trace to x=1 to find the vertex at (1, 18) and hands fly up, eager lips whisper to each other, 3 +/- i sqrt(18), and nods in return assure them they are right... so the teacher springs his trap... pointing, a student responds with certainty... and the teacher tilts his head, and gives him "the eye".... "Did you check?"

A timid young lady offers, "I got an answer of 1 +/- 3i." Heads lean toward each other. What happened. I was sure we had it... and then, an offering of insight... "It's half as much, I mean, its the square root of half of the y part...because of the two in front." But it is almost a question. The bell rings, and no one moves... "Is that it? Tell us."

He turns to the board to hide his smile as he erases.... painfully slowly in their mind..then turns back.... and shrugs... then offers... "Hey, have a great weekend."

Pick Two

A colleague from Colorado sent me an interesting probability problem the other day. I like it because it illustrates one of those serendipitous qualities of mathematics. Here is the problem. A Jar has a mixture of Red and White balls so that if you withdraw two, the probability of getting two alike, or two of different color are both equal to one-half. You may want to stop and try it before you read on.

Ok, so we let r = the number of red ones and W be the number of white ones, and the total is r+w. So how could we draw one of each color? Well, red first, and then white, or white first and then red. If we find the probability of each of these conditional events and add them up, that will have to equal 1/2. Ok, the probability of red on the first draw is r/(r+w), and on the second ball the probability is w/(r+w-1) since one of the red balls will be missing. The opposite order is exactly the same with the w and r reversed, so the probability of getting one of each color is 2rw/[(r+w)(r+w-1)]. Setting that equal to 1/2 we get

If we exapand (r+w)² and subtract the 4rw we get 0=r²-2rw +w²-r-w. NOTICE the symmetry, we could exchange r and w and get the same equation. We know right off that any solution (a,b) will have another soltuion (b,a).

One of the things that is often hard for students is to think of one variable as a constant and the other as a variable. I like to use the word "pronumeral", like a pronoun only instead of him or her we say "that number". It is like a variable that doesn't vary, we just don't know what it is in a particular case. So think of w as if it were fixed. We have that many white balls in the jar and we are wondering how many red can be put in to make the problem work... see it.. w is a "fixed" unknown, but r is going to "vary". That makes the equation a quadratic in r; Ar² +Br+c=0 where A=1, B= -2w-1, and C=w²-w.

We can solve this using the quadratic formula, but if this solution is going to be a rational number, and the number of balls in a jar must be rational, then the discriminant, the expression under the squre root radical in the quadratic forula, B²-4AC, must be a perfect square. B²= 4w²+4w+1 and 4AC= 4w²-4w; so B²-4AC= 8W+1. If there is a rational solution, it must be when 8W+1 is a perfect square. Wait, I know this one! That's a problem from number theory. The numbers that make 8W+1 a perfect square are called triangular numbers; 1, 3, 6, 10, 15. They are the sum of the first n counting numbers. But a neat thing happens if we plug 1 in for W, the solution for r is 3.... and if we use 3 for w, r=6. Each time we substitute one of the triangular numbers into the quadratic, the next comes out as a solution. So the probability of drawing two balls of the same color, (or of two that are not alike) will equal 1/2 whenever the number of balls of each color are consecutrive triangular numbers. A very geometric solution to a very algebraic question.