Algebra for Dopes - It Ain't That Hard - The Quadratic Equation - Part III

In Part III of this series, we examine how we solve factorable quadratics of the form ax^2 - bx + c, in which both the a and c terms are positive, but the b, or middle term, is negative. Such an example would be x^2 - 5x + 6. This class of quadratics, which we will call the "b-negatives" are easily solvable using the factoring techniques we learned about in Part II.

To get right to the chase, let's look at the quadratic in the first paragraph: x^2 -5x + 6. Remember, we are looking for the solutions, or those values of x which when plugged in for this variable make the quadratic zero. This is another way of saying that those values are the zeros of the function, or those values are the ones which make the quadratic vanish. As in Part II, we look to see the factors of the c-term, which in this example is 6. They are: 1-6, and 2-3. Now this is a very good example because of the nature of these two factor pairs. You see, both 1-6 and 2-3 can combine in some way to give 5; for 1 and 6 can be combined subtractively, that is 6-1 = 5; and 2 and 3 can be combined additively, that is 2 + 3 = 5.

To decide which pair works, remember that this article treats the b-negative class of quadratics. Since the c-term, or 6 is obtained from multiplying the factor pair and only two positives or two negatives can produce a positive result, and since the b-term or -5 can only be obtained by adding two negative numbers, the only factor pair that works is 2-3. For if it were 1-6, then -1 and -6 add to -7 not -5. Thus x^2 - 5x + 6 = (x - 3)(x - 2), and the solutions are 3 and 2. The reader can easily verify that either of these values substituted for x in the original quadratic makes the equation vanish.

To see how this method works with one more example, let's try x^2 - 10x + 16. The factor pairs of 16 are 1-16, 2-8, and 4-4. The only pair that will combine to give 10 is 2-8. Since we have a b-negative quadratic we must have both of these numbers as negatives. Therefore, the quadratic factors into (x - 8)(x - 2) and the solutions are 8 and 2. Nothing at all difficult about this.

If you have followed this series, you are starting to see how we take algebra down to size. By "chunking" this subject up into little component pieces, we make it a much more manageable beast. Thus far we have dissected the class of factorable quadratics into two subclasses, each of which is very easy to deal with. Once this view is taken, the rest of algebra becomes just as manageable and much more pleasant.