As mentioned in Part I of this series, mastering algebra requires little more than clear-cut explanation and some concerted practice. Conquering the diverse aspects of this field necessitates focused thought and willingness to see through the trees to view the forest. Rather than view algebra-which encompasses many different topics and has vast extensions throughout-in one fell swoop, it is much better to view this subject piecemeal. This is the approach that will be taken in this series of articles.
Learning quadratic equations is one of the key plateaus in algebra as this opens the door to higher degree equations and even the calculus. One of the difficulties with this subject area is that the quadratic has many ways of being conquered, so to speak, that is to say, solved. Essentially there are three ways to solve a second-degree, or quadratic equation: one is by factoring; two is by completing the square; and three is by using the famous quadratic formula.
Now this is where it gets a little sticky since each of these three methods is a whole field within itself. Of the three methods, both the formula and the completing the square will always work on any quadratic equation. Factoring will only work on a subclass of the entire class of second-degree equations, and it is this class that we will discuss briefly here. Specifically, we will look at the subclass of factorable quadratics which have terms with all positive coefficients. That is, if we consider the general quadratic ax^2 + bx + c, then a, b, and c will all be positive. For the sake of classification, we will call this subclass of quadratics the "abc positives."
In order to solve an "abc positive" by factoring, it is necessary that the multiplicative factors of the c-term be able to combine additively or subtractively to the b-term. Now that this is starting to sound like a lot of mumbo jumbo, let's break this down and show how simple this really is. Let's do this with a specific example. Take x^2 + 7x + 12. The pair-wise factors of 12 are 1 and 12, 2 and 6, and 3 and 4. When we multiply any of these pair-wise factors we get 12. We look for the pair that combines additively or subtractively to give 7. It is obvious that only 3 and 4 do.
From this, we factor as such: x^2 + 7x + 12 = (x + 3)(x + 4). The solution to this quadratic is now the opposite of 3 and 4, or -3 and -4. That is, if we substitute -3 or -4 into the original quadratic equation, the result will be 0, for this is what it means for a number to solve a quadratic equation: it makes the equation 0 upon substitution. I leave it to the reader to verify that -3 and -4 do indeed do this.
After taking a look at one more, you will now have the secret to solving the entire subclass of abc positives. Take x^2 + 9x + 20. The pair-wise factors of 20 are 1-20, 2-10, and 4-5. The only pair that add to 9 are 4-5. Thus x^2 + 9x + 20 = (x + 4)(x + 5) and the solutions are -4 and -5. Once again, you can confirm that when each of these values is substituted back into the original quadratic, you will get 0.
When the vast subject of algebra is broken down to size, you start to see that it really is not as difficult as some might make it. In fact, you might start to see that it really is quite understandable. After looking at this approach, you come to realize that anyone could master this field. It really is true: algebra really ain't that bad. And once algebra is mastered, you can be thrilled to know that mathematics opens up a magical door which leads to a rich world of discovery.
0 comments:
Post a Comment