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

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.

See more at Cool Math Site and Cool Algebra Ebooks

Joe is a prolific writer of self-help and educational material and an award-winning former teacher of both college and high school mathematics. Under the penname, JC Page, Joe authored Arithmetic Magic, the little classic on the ABC's of arithmetic. Joe is also author of the charming self-help ebook, Making a Good Impression Every Time: The Secret to Instant Popularity; the original collection of poetry, Poems for the Mathematically Insecure, and the short but highly effective fraction troubleshooter Fractions for the Faint of Heart. The diverse genre of his writings (novel, short story, essay, script, and poetry)-particularly in regard to its educational flavor- continues to captivate readers and to earn him recognition.

Joe propagates his teaching philosophy through his articles and books and is dedicated.

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.

See more at Cool Math Site and Cool Algebra Ebooks

Joe is a prolific writer of self-help and educational material and an award-winning former teacher of both college and high school mathematics. Under the penname, JC Page, Joe authored Arithmetic Magic, the little classic on the ABC's of arithmetic. Joe is also author of the charming self-help ebook, Making a Good Impression Every Time: The Secret to Instant Popularity; the original collection of poetry, Poems for the Mathematically Insecure, and the short but highly effective fraction troubleshooter Fractions for the Faint of Heart. The diverse genre of his writings (novel, short story, essay, script, and poetry)-particularly in regard to its educational flavor- continues to captivate readers and to earn him recognition.

Joe propagates his teaching philosophy through his articles and books and is dedicated.

Algebra For Dopes - It Ain't That Hard - The Quadratic Equation - Part V

In Part V of this series, we examine how we solve the last class of factorable quadratics of the form ax^2 + bx - c, in which the b-term is positive and the c-term is negative. Such an example would be x^2 + 4x - 5. This subclass of quadratics are as easily solvable as those of the "bc-negative" class discussed in Part IV of this series.

To show how similar this class is, let's examine x^2 + 4x -5. This is the same quadratic as the first example in Part IV of this series, except the 4x term now is positive instead of negative. As in the last article, we note that 5 is prime and its only factors are 1 and 5. Since the c-term is negative, the 1 and 5 must be of opposite signs. Since the b-term is positive, the larger number must bear the positive sign; otherwise the result of the b-term would be negative. Thus x^2 + 4x - 5 = (x + 5)(x - 1), and the solutions are -5 and 1.

Although this method should be perfectly clear by now, let's reinforce it with two more examples. Let's take the quadratic x^2 + 10x - 24. The factor pairs of 24 are 1-24, 2-12, 3-8, and 4-6. Notice that as the c-term becomes a larger composite number as in this case, generally the number of possible factor pairs increases. When the c-term is a prime number, as in the first example, then the only factor pairs are 1 and the number itself. (By the way the first 10 primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29; 2 is the one and only even prime; 1 is not considered a prime number.) Now of the four factor pairs, the only one that can combine to give 10 is 2-12. Since 10 must be positive and 2 and 12 must be of opposite sign, can you guess which must be positive and which negative? Easy enough, right? Thus x^2 + 10x - 24 = (x + 12)(x - 2) and the solutions are -12 and 2.

Finally, we will solve x^2 + 31x - 66. The factor pairs of 66 are 1-66, 2-33, 3-22, and 6-11. The only pair that combines to yield 31 is 2-33. Again, using the argument above this quadratic must factor as (x + 33)(x - 2), and the solutions are -33 and 2. The reader can easily verify that both -33 and 2 are in fact the zeros of this particular quadratic.

After following this series of articles, you are starting to see how quick you can become at algebra once you understand the rules of the game. And this goes for all of algebra: as we break down each component of this subject and apply these techniques, algebra-and indeed math-no longer is a mystery that perplexes, but a mystery that both enriches and enlightens.

See more at Cool Math Site and Cool Algebra Ebooks

Joe is a prolific writer of self-help and educational material and an award-winning former teacher of both college and high school mathematics. Under the penname, JC Page, Joe authored Arithmetic Magic, the little classic on the ABC's of arithmetic. Joe is also author of the charming self-help ebook, Making a Good Impression Every Time: The Secret to Instant Popularity; the original collection of poetry, Poems for the Mathematically Insecure, and the short but highly effective fraction troubleshooter Fractions for the Faint of Heart. The diverse genre of his writings (novel, short story, essay, script, and poetry)-particularly in regard to its educational flavor- continues to captivate readers and to earn him recognition.

Joe propagates his teaching philosophy through his articles and books and is dedicated to helping educate children living in impoverished countries. Toward this end, he donates a portion of the proceeds from the sale of every ebook. For more information go to http://www.mathbyjoe.com

Algebra For Dopes - It Ain't That Hard - The Quadratic Equation - Part IV

In Part IV of this series, we examine how we solve factorable quadratics of the form ax^2 - bx - c, in which both the b and c terms are negative. Such an example would be x^2 - 4x - 5. This subclass of quadratics, which we will call the "bc-negatives," are easily solvable using the factoring techniques we learned about in Parts II and III of this series of articles.

Once again, let's get right to the point and solve the quadratic in the first paragraph: x^2 -4x -5. 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 III, we look to see the factors of the c-term, which in this example is -5. When considering the factor pairs, we will disregard the negative sign and only examine the factors of 5. Since 5 is a prime number, its only factors are 1 and itself.

Remember: the c-term of the quadratic is always formed by multiplying the numbers of the factor pair. Since the c-term is negative, the 1 and 5 must be of opposite sign, since this is the only way we can get a negative product. Thus we must have 1 with a -5 or -1 with a 5. Both (1)(-5) and (-1)(5) = -5. Since the b-term is obtained by adding and since the b-term is negative, the larger number of the factor pair must be negative. If this were not true, then the result would be positive. You can think of this as the larger negative force overpowering the smaller positive force to give a net negative force. You can also relate this back to chemistry in that if you have more negative electrons than positive protons you end up with a net negative charge, and vice versa. (You see how you can relate math back to many other things in the real world.) Consequently, this quadratic must factor as (x - 5)(x + 1) and the solutions are 5 and -1. Once again, you can plug either of these two values back into the original quadratic to show that they make the equation equal to zero.

To see how this method works with another example, let's try x^2 - 4x - 12. The factor pairs of 12 are 1-12, 2-6, and 3-4. The only pair that will combine to give 4 is 2-6. Now we have to make an adjustment for the b-term. We know that (2)(-6) and (-2)(6) = -12. Of these two examples, the greater negative force occurs in the pair (2)(-6) to yield a net negative of -4. This is what we need, and hence x^2 - 4x - 12 = (x + 2)(x - 6), and the answers are -2 and 6. Each of these will make the quadratic vanish or become equal to zero upon substitution.

In the next article, we will look at the final case of factorable quadratics, ones of the form

x^2 + bx - c. With all the tools we now have available, we will be able to dispatch with this last class with the blink of an eye. In fact, if you've followed this series this far, you already have the knowledge to do so now. Stay tuned...

See more at Cool Math Site and Cool Algebra Ebooks

Joe is a prolific writer of self-help and educational material and an award-winning former teacher of both college and high school mathematics. Under the penname, JC Page, Joe authored Arithmetic Magic, the little classic on the ABC's of arithmetic. Joe is also author of the charming self-help ebook, Making a Good Impression Every Time: The Secret to Instant Popularity; the original collection of poetry, Poems for the Mathematically Insecure, and the short but highly effective fraction troubleshooter Fractions for the Faint of Heart. The diverse genre of his writings (novel, short story, essay, script, and poetry)-particularly in regard to its educational flavor- continues to captivate readers and to earn him recognition.

Joe propagates his teaching philosophy through his articles and books and is dedicated to helping educate children living in impoverished countries. Toward this end, he donates a portion of the proceeds from the sale of every ebook. For more information go to http://www.mathbyjoe.com

Algebra For Dopes - It Ain't That Hard - The Quadratic Equation - Part V

In Part V of this series, we examine how we solve the last class of factorable quadratics of the form ax^2 + bx - c, in which the b-term is positive and the c-term is negative. Such an example would be x^2 + 4x - 5. This subclass of quadratics are as easily solvable as those of the "bc-negative" class discussed in Part IV of this series.

To show how similar this class is, let's examine x^2 + 4x -5. This is the same quadratic as the first example in Part IV of this series, except the 4x term now is positive instead of negative. As in the last article, we note that 5 is prime and its only factors are 1 and 5. Since the c-term is negative, the 1 and 5 must be of opposite signs. Since the b-term is positive, the larger number must bear the positive sign; otherwise the result of the b-term would be negative. Thus x^2 + 4x - 5 = (x + 5)(x - 1), and the solutions are -5 and 1.

Although this method should be perfectly clear by now, let's reinforce it with two more examples. Let's take the quadratic x^2 + 10x - 24. The factor pairs of 24 are 1-24, 2-12, 3-8, and 4-6. Notice that as the c-term becomes a larger composite number as in this case, generally the number of possible factor pairs increases. When the c-term is a prime number, as in the first example, then the only factor pairs are 1 and the number itself. (By the way the first 10 primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29; 2 is the one and only even prime; 1 is not considered a prime number.) Now of the four factor pairs, the only one that can combine to give 10 is 2-12. Since 10 must be positive and 2 and 12 must be of opposite sign, can you guess which must be positive and which negative? Easy enough, right? Thus x^2 + 10x - 24 = (x + 12)(x - 2) and the solutions are -12 and 2.

Finally, we will solve x^2 + 31x - 66. The factor pairs of 66 are 1-66, 2-33, 3-22, and 6-11. The only pair that combines to yield 31 is 2-33. Again, using the argument above this quadratic must factor as (x + 33)(x - 2), and the solutions are -33 and 2. The reader can easily verify that both -33 and 2 are in fact the zeros of this particular quadratic.

After following this series of articles, you are starting to see how quick you can become at algebra once you understand the rules of the game. And this goes for all of algebra: as we break down each component of this subject and apply these techniques, algebra-and indeed math-no longer is a mystery that perplexes, but a mystery that both enriches and enlightens.

See more at Cool Math Site and Cool Algebra Ebooks

Algebra For Dopes - It Ain't That Hard - The Quadratic Equation - Part IV

In Part IV of this series, we examine how we solve factorable quadratics of the form ax^2 - bx - c, in which both the b and c terms are negative. Such an example would be x^2 - 4x - 5. This subclass of quadratics, which we will call the "bc-negatives," are easily solvable using the factoring techniques we learned about in Parts II and III of this series of articles.

Once again, let's get right to the point and solve the quadratic in the first paragraph: x^2 -4x -5. 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 III, we look to see the factors of the c-term, which in this example is -5. When considering the factor pairs, we will disregard the negative sign and only examine the factors of 5. Since 5 is a prime number, its only factors are 1 and itself.

Remember: the c-term of the quadratic is always formed by multiplying the numbers of the factor pair. Since the c-term is negative, the 1 and 5 must be of opposite sign, since this is the only way we can get a negative product. Thus we must have 1 with a -5 or -1 with a 5. Both (1)(-5) and (-1)(5) = -5. Since the b-term is obtained by adding and since the b-term is negative, the larger number of the factor pair must be negative. If this were not true, then the result would be positive. You can think of this as the larger negative force overpowering the smaller positive force to give a net negative force. You can also relate this back to chemistry in that if you have more negative electrons than positive protons you end up with a net negative charge, and vice versa. (You see how you can relate math back to many other things in the real world.) Consequently, this quadratic must factor as (x - 5)(x + 1) and the solutions are 5 and -1. Once again, you can plug either of these two values back into the original quadratic to show that they make the equation equal to zero.

To see how this method works with another example, let's try x^2 - 4x - 12. The factor pairs of 12 are 1-12, 2-6, and 3-4. The only pair that will combine to give 4 is 2-6. Now we have to make an adjustment for the b-term. We know that (2)(-6) and (-2)(6) = -12. Of these two examples, the greater negative force occurs in the pair (2)(-6) to yield a net negative of -4. This is what we need, and hence x^2 - 4x - 12 = (x + 2)(x - 6), and the answers are -2 and 6. Each of these will make the quadratic vanish or become equal to zero upon substitution.

In the next article, we will look at the final case of factorable quadratics, ones of the form

x^2 + bx - c. With all the tools we now have available, we will be able to dispatch with this last class with the blink of an eye. In fact, if you've followed this series this far, you already have the knowledge to do so now. Stay tuned...

See more at Cool Math Site and Cool Algebra Ebooks

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

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.

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.