The sum - product problem

Question

I have long been a fan of all of the different methods for factoring quadratics, yet I hardly ever use them in my classroom. The first task they are confronted with, in factoring trinomials, is to find two numbers whos sum is this and whos product is that. I call this the sum-product problem.

Usually I teach my students to solve the sum-product problem by having them list all of the pairs of factors. The drawback with this is that students have trouble with what should be positive or negative and do they add or subtract.

I came up with this method recently and I like it because it's deterministic but I wonder about the amount of calculation involved. I'd like to know what plusses and minuses you see in it. I will give you a worst-case example.

EXAMPLE: Find two numbers that add to -7 and multiply to -60.

Half of -7 is $-\frac 7 2$.

So the two numbers must be $-\frac 7 2 - x$ and $-\frac 7 2 + x$ for some positive $x$.

Thus we need to solve $(-\frac 7 2 - x) (-\frac 7 2 + x) = -60$

$\frac{49}{4} - x^2 = -60$

$x^2 = \frac{49}{4} + 60 = \frac{289}{4}$

$x = \frac{17}{2}$

So the two numbers are

$-\frac{7}{2} - \frac{17}{2} = -\frac{24}{2} = -12$

and

$-\frac{7}{2} + \frac{17}{2} = \frac{10}{2} = 5$

Answer 1

The method is marvelously clever and straightforward and I can see people thinking it's very cool (which it is).

I do feel it would be clearer if it began like this: "Let $x$ and $y$ be the two numbers. Since they sum to $-7$, their average is $-7/2$ and so one is more than $-7/2$ and the other is less. Write $x=-7/2 + d$ and $y = -7/2 - d$." Then you're explicitly declaring that your real goal is the two numbers and this other number is just a means to an end.

The idea that if two numbers add to $A$, then one must be a bit more than $A/2$ and the other must be a bit less than $A/2$ ... I feel this idea would be quite a conceptual leap for many students. It's certainly something that it is worth them knowing, but I can bet quite a few of them will be very confused by that particular step.

There's a little nuance in there that the number $x$ you have chosen is implicitly positive. Students familiar with putting a $\pm$ when they perform a square root will get two answers for $x$ and sub them both back in to get four answers (which turn out to only be two answers after all). This will waste quite a bit of time unless you explicitly point out that the $x$ you're using is positive.

The calculations involved are actually exactly the same as those you do when completing the square: divide the x-coefficient by 2, square it, add this to the constant, square root, add and subtract this to minus half the x-coefficient. So in terms of calculations it is actually equivalent to completing the square. Completing the square feels harder for some reason to me -- I'm not sure why. Still, completing the square is a method they probably need to know for general quadratics, so is it worth showing them something else for a particular problem?

The major problem to me is that the reasoning in it only applies to this specific sort of problem, and so might encourage the "list of types of problems - list of solution methods" approach to maths. I can't see any other context in which this type of reasoning would come up, so it seems to me that you'll be teaching them to say to themselves "there's a clever trick to these ones, let's see if I can remember it" rather than "If I just use plain ordinary reasoning and common general methods I'm sure I can figure it out."

Finally, there are actually thought processes that will tell you precisely whether the factors have to be positive or negative (and you always add them since it says "sum"). If the product is positive then it must be that both factors are positive or both are negative. Then if the sum is positive this is only possible if both factors are positive, and if the sum is negative this is only possible if both factors are negative. On the other hand if the product is negative then it must be that one factor is negative and one is positive. Then if the sum is negative it must be that the negative factor is bigger in size than the positive one, and if the sum is positive then it must be that the negative factor is smaller in size than the positive one. So in short:

            SUM +                          SUM -
PRODUCT +   both factors +                 both factors -
PRODUCT -   big factor +, small factor -   big factor -, small factor +

In your example, the pairs of factors of 60 are:
1, 60; 2, 30; 3,20; 4,15; 5,12; 6,10.
Since the product is negative we have to have one + and one -. Since the sum is negative the big one has to be -ve. So the pairs of factors we need are actually:
1, -60; 2, -30; 3, -20; 4, -15; 5, -12; 6, -10.
So it must be 5 and -12.

Answer 2

It is likely too early to show them this...?

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$$x+y = -7, \;\; \text{and} \; \; x y = -60 \;.$$

Answer 3

This idea can actually be used to solve any quadratic equation, and in fact to obtain the quadratic formula, but for some reason it never seems to get mentioned in textbooks, even those that lay all the appropriate groundwork by spending a lot of time with "easily solved quadratics", such as $(x+5)^2 - 6 = 30.$ What follows is a post at Math Forum I made fairly recently (25 November 2014), in which the only changes I've made have been to convert the math symbols to LaTeX form.

Here's a way that can be motivated from $x^2 = c$ $(c \geq 0)$ and factoring methods. One would have already observed that $x^2 = c$ can be rewritten as $(x - \sqrt {c})(x + \sqrt {c}) = 0,$ which gives rise to the solutions $x = \pm \sqrt {c}.$

In general, quadratic equations having no linear term are easy to solve, and students should have had plenty of practice with such equations before what follows, including equations such as $(x+5)^2 - 6 = 30.$

In considering $x^2 + bx = c$ (any quadratic equation can be rewritten this way without altering the solution set), some students want to consider $x(x+b) = c.$ This is not wrong! What is wrong is to continue and try to extend the zero property to the possibly nonzero number $c.$

In considering $x(x+b) = c,$ you might point out it's a shame this isn't something like $(x-d)(x+d) = c,$ because when expanding a difference of squares the linear term drops out, and then the equation would be easy to solve.

Let's look closer at what's not working for a specific case.

Suppose we want to solve $x(x+6) = 16.$

Now $x(x+6)$ can be written as $(x-0)(x+6),$ which is half-way to being a difference of squares. What messes things up is that $0$ and $6$ are not the same number. The lightbulb idea is to notice that if were dealing with $x+3$ instead of $x,$ then the factors $x-0$ and $x+6$ are exactly $3$ left and $3$ right of $x+3.$ In other words, instead of trying to solve for $x,$ instead consider the problem of solving for what lies midway between $x$ and $x+6,$ which is $x+3.$ And note that if we can find the values for $x+3,$ then we can just subtract $3$ to get the values for $x.$

$$x(x+6) \;\; = \;\; [(x+3) - 3][(x+3) + 3] \;\; = \;\; 16$$

Expanding (not completely, but in a smart way) gives

$$(x+3)^2 - 9 \; = \; 16$$ $$(x+3)^2 \; = \; 25$$

$$x+3 \; = \; -5, \; 5$$

$$x \; = \; -5 - 3, \; 5 - 3$$

$$x \; = \; -8, \; 2$$

Doing the same thing algebraically to $x^2 + bx = c$ gives the quadratic formula for equations of the form $x^2 + bx - c = 0.$

(comments not included in my original post) Some readers will recognize that this is essentially the same technique as the standard variable substitution that gets rid of the linear term, namely $u = x + \frac{b}{2}.$ Note that this is simply putting $u$ equal to the average of $(x-r_{1})$ and $(x-r_{2}),$ where $r_1$ and $r_2$ are the two solutions, since

$$\frac{(x - r_{1}) \; + \; (x - r_{2})}{2} \;\; = \;\; x - \frac{r_1 + r_{2}}{2} \;\; = \;\; x + \frac{b}{2},$$

where the last equality follows from $r_1 + r_2 = -b$ (one of the Vieta formulas). The same idea is used to get rid of the quadratic term in a cubic equation, namely put $u$ equal to the average of $(x-r_{1})$ and $(x-r_{2})$ and $(x-r_{3}),$ where $r_{1},$ $r_{2},$ $r_{3}$ are the three solutions, and then make use of the Vieta formula for the sum of the roots of a cubic equation in order to get the substitution explicitly in terms of the coefficients of the cubic equation (rather than explicitly in terms of the solutions).

Answer 4

A little bit of historical context: This is, in fact, precisely the method demonstrated by Diophantus of Alexandria in The Arithmetica. In Book I, Problem 27 states:

27. To find two numbers such that their sum and product are given numbers.

Necessary condition. The square of half the sum must exceed the product by a square number...

Given sum 20, given product 96.

$2x$ the difference of the required numbers. Therefore the numbers are $10+x$, $10-x$. Hence $100-x^2 = 96$. Therefore $x=2$, and the desired numbers are $12, 8$.

The above is from p. 140 of the translation of Heath (2nd edition, 1910), which may be downloaded from the Internet Archive.

The "necessary condition" is necessary only if you are looking for whole-number solutions, of course. If we are okay using irrational numbers, it is enough to have the weaker condition that the square of half the sum must be greater than or equal to the product.

Answer 5

Mostly I agree with @DavidButlerUofA had said in his answer. Clear up some of the language to make it more sound and it's a pretty good method for factoring quadratics.

It is limited by the fact that it isn't easy to apply in other contexts, but neither are many of the other techniques students use to factor quadratics. My suggestion to cope with that is to frame this in the context of the variety of techniques for factoring quadratics. When I teach this topic I try to explore a variety of strategies to factor and solve quadratics, highlighting graphing and completing the square. After some exploration we usually settle into one or two strategies for the class and that's what I'll use in class for explanations. My thought is that it helps to establish that we're choosing to use a particular strategy, and I hope it limits the harm done when students believe there is a "right way" to solve a problem.

Answer 6

It seems to me that if you are going so far as to invent your own specialized completing-the-square-like method, you might as well use a widely-accepted completing-the-square-like method:

To factor $x^2 - 7x - 60$, we can use the quadratic formula to find the roots of $x^2 - 7x - 60 = 0$.

$$\frac{7 \pm \sqrt{289}}{2} = 12 \text{ and } -5$$

So $x^2 - 7x - 60 = (x-12)(x+5)$.

Answer 7

What you done is Viete, never lieked that. I admit to like your way of putting it but... I much more prone to use this:

1. step 1: get to standart form aX^2 + bX + c = 0;
2. step 2: d = b^2 -4ac;
3. step 3: X1 = (-b-sqrt(d))/2a; X2 = (-b+sqrt(d))/2a;

Imho that is way easier

Viete would be X1 + X2 = -b/a; X1*X2 = c/a;