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.

A B C of Solving Quadratic Equation

An expression of the type a x ² + b x + c = 0 , ( a ≠ 0 ) is called a quadratic equation in the variable x .

The equation a x ² + b x + c = 0 is called the general (or, standard form)

We can solve a quadratic equation by (1) factorization or by (2) applying the formula.

The formula of finding the roots of the quadratic equation is as follows

x = (- b ± √ (b ² - 4 a c) ) / 2 a

Now we will discuss how to solve applied problems. Due to wide variety of applied problems, there is no single solving technique that works in all cases. However the following suggestion proved helpful.

Step: 1 Read the problem carefully and determine what quantity (s) must be found.

Step: 2 assign a variable name to the quantity.

Step: 3 try expressing the problem algebraically, and as well determining which expressions are equal and write the necessary equation (s).

Step: 3 solve the resulting equation (s)

Now go through a simple problem based on formation of quadratic equation and solving

Problem: The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 58 / 21 . Find the fraction.

Solution:

Let the numerator of the fraction be x (x be an element of I )

Then its denominator is (2x + 1).

So the fraction is x / (2x + 1))

And the reciprocal would be (2 x + 1) / x

According to the problem:

(X / (2 xs + 1)) + ((2 x + 1) / x) = ( 58 / 21 )

(X ² + (2 x + 1) ²) / (x + (2x + 1)) = ( 58 / 21)

[L. C. D is = ( x + ( 2x + 1 ) ]

21 ( x ² + 4 x ² + 4 x + 1 ) = 58 x ( 2x + 1 )

105 x² + 84 x + 21 = 116 x ² + 58 x

11 x ² - 26 x - 21 = 0

11 x ² - (33 - 7) x - 21 = 0 [using middle term factorization]

11 x ²- 33 x + 7 x - 21 = 0

11 x ( x - 3 ) + 7 ( x - 3 ) = 0

(x - 3) (11 x + 7) = 0

Either ( x - 3 ) = 0 , or ( 11x + 7 ) = 0 [ using zero factor theorem ]

x = 3 ,

From, (11x + 7) = 0

We get, x = - (7 / 11)

But x is an integer , neglect x = - (7 / 11)

Take x = 3

So the required fraction, (x / (2x + 1)) = (3 / (2 * 3 + 1))

= (3 / 7)

Now try the following:

The age of a man is twice the square of the age of his son .Eight years hence the age of the man will be 4 years more than thrice the age of his son. Find their present age?

If you cannot solve this problem, you probably need more practice. A good online tutor would be helpful if you plan to master this subject in a short period. Any reasonably good math tutor should do.

Ishani Dutta is an educator who specializes in providing online tuition packages for Mathematics and English. For similar tips on different topics like pre-algebra, co-ordinate geometry, essay writing,english grammar, mailto:info@learningexpress.biz or go to: [http://www.learningexpress.biz] for online help

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

Mastering Algebra - The Equation of the Circle

As one of the conic sections, the circle is probably the most important of these curves. When studying analytic geometry (the relationship between the algebraic formula for a curve and the actual graph) students are required to learn how to recognize the circle as well as to graph it. Here we discuss the simplest way to recognize this curve, put it into suitable algebraic form, and graph it on a coordinate grid.

The formal definition of the circle is the locus, or site, of points that are all equidistant from another fixed point. The set of points forming the circle outline its circumference; the fixed point is the center. The distance from the fixed point, or center, to any point on the periphery of the circle is the radius.

When in standard form, the equation of the circle takes on the following form: (x - h)^2 + (y - k)^2 = r^2. The center is located at the point (h,k) and the radius is r. Once we get the algebraic equation into such form, graphing could not be easier, as we simply plot the point (h,k) on our grid, and then go r units from this point up, down, to the left, and to the right. We then do our best to connect these points by a smooth circle.

To put the equation into standard form often requires a technique known as completing the square. As in life, the things we usually need require some work to get and this is no different in mathematics. Most equations are not so neat and tidy so as to be in standard form at first blush; therefore, we need to manipulate the equation a bit to get it into good form. This is not difficult however, and we shall show by example how this is done. Once in standard form, the center and radius are obvious and the graph becomes readily accessible.

Take the equation x^2 + y^2 + 2x + 4y - 4 = 0. This is obviously not in the form (x - h)^2 + (y - k)^2 = r^2. However, with a little manipulation, we can put this into such form. This procedure works no matter what the equation, as long as the equation is that of a circle. The only things that change are the numbers. Thus once you follow this procedure, you can put any equation which will produce a circle into standard form.

First isolate both the x and y terms and write as such: x^2 + 2x + y^2 + 4y - 4 = 0. Now bring the -4 over to the right side, and write as such: x^2 + 2x + y^2 + 4y = 4. We now complete the square on x and y by taking half of the coefficient of each and squaring both terms. Half of 2 is 1 and half of 4 is 2. Squaring each of these terms give 1 and 4, respectively, and adding them to both sides of the equation results in x^2 + 2x + 1 + y^2 + 4y + 4 = 4 + 5 = 9. Now we have two perfect square trinomials in x and y. These are always factorable into a form which puts both the x and y terms into standard form for the equation of the circle. The x^2 + 2x + 1 becomes (x + 1)^2 and the y^2 + 4y + 4 becomes (y + 2)^2. Notice that the h and k are -1 and -2, the opposite of what is inside parentheses. Notice also that the 1 and 2 are the terms which were derived by halving the coefficients of the x and y terms.

Thus we have x^2 + y^2 + 2x + 4y - 4 = 0 becomes (x + 1)^2 + (y + 2)^2 = 9. Observe that 9 is 3^2. Consequently, we have (x + 1)^2 + (y + 2)^2 = 3^2. Looking at this equation, we see that the center is (-1, -2) and the radius is 3. From this equation, we plot the center and move 3 units up, down, left, and right. We then draw a smooth curve. This procedure is exactly the same for every circle equation. The only things that change are the numbers.

You now have the tools to slay any circle equation or graph. Just follow the simple procedure above and you will be able to conquer any algebra problem that involves putting circle equations into standard form and graphing. After all, you probably have many other things to put your attention to, such as getting that new iPhone. Now you don't have to worry about circles any more. Enjoy.

High School Algebra Finally Pays Off - The Body Fat Equation

When approaching your goals for a better you, every game plan needs to start with determining where you are, and where you want to end up. We're going to talk about some ways to assess where you are -- right now -- so that you can make sure that the goals you have set (or will set) are reasonable. And yes, this will be on the test!

Whenever going into battle, a commander will send out reconnaissance experts to collect information on the size and strength of the enemy. We will continue to use a "battle" theme in this article, because your fight to lose weight and lose fat is nothing short of that.

Just as the commander needs to have a good assessment of what he or she is facing, we need to have an accurate assessment of our current situation so that we can set proper and appropriate goals. An inappropriate goal can insure failure, or worse, cause us physical harm.

If your goal is to look like a swimsuit model in time for your Hawaiian vacation a month and a half from now, you'd better not need to lose more than a few pounds. If your goal is to lose ten pounds per week for the next six weeks, you're simply not being practical.

With that said, what is 'reasonable'? You probably know how much you weigh, but do you know what percentage of your overall weight is fat?

The best way to know is to see a doctor or go to a gym that can calculate your body fat percentage. Normal body fat levels for women are between 22 - 25% and between 15 - 18% for men. Lean people will be slightly less, and athletes even lower. In lieu of that, there is a simple calculation that serves as a good guide, but is no where near as accurate.

This method is called the BMI or Body Mass Index and was created by the National Institutes of Health in 1998. The BMI uses your height and your weight to generate a factor that suggests whether you are underweight, normal, overweight or obese.

The BMI is generalized, so it cannot take frame size into account. If you have a large frame, your BMI will seem worse than it is, if you have a small frame, your BMI will seem better than it is. Use it as a guide and then add a portion of common sense.

In order to calculate your BMI, use the following equation:

BW = Body Weight in Pounds

HI = Height in Inches

(BW / (HI * HI) ) * 703

You will need a calculator to solve this equation.

If you failed algebra, here is the equation in English:

Divide your body weight in pounds (BW) by the square* of your height in inches (HI) and multiply that result by 703.

Once you've calculated your BMI, use the following guidelines to find out how you measure up:

BMI 18.5 AND 24.9 AND 30 You are in the "Obese" zone.

Again, this gives you a rough guide, but it also gives you (possibly) a more realistic view of your current situation.

This more realistic idea will help you to set more realistic and achievable goals, which lead to a greater degree of success!

( * - The "SQUARE" of a number means the number times itself. The "square" of 4 is 16.)

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.

A B C of Solving Quadratic Equation

An expression of the type a x ² + b x + c = 0 , ( a ≠ 0 ) is called a quadratic equation in the variable x .

The equation a x ² + b x + c = 0 is called the general (or, standard form)

We can solve a quadratic equation by (1) factorization or by (2) applying the formula.

The formula of finding the roots of the quadratic equation is as follows

x = (- b ± √ (b ² - 4 a c) ) / 2 a

Now we will discuss how to solve applied problems. Due to wide variety of applied problems, there is no single solving technique that works in all cases. However the following suggestion proved helpful.

Step: 1 Read the problem carefully and determine what quantity (s) must be found.

Step: 2 assign a variable name to the quantity.

Step: 3 try expressing the problem algebraically, and as well determining which expressions are equal and write the necessary equation (s).

Step: 3 solve the resulting equation (s)

Now go through a simple problem based on formation of quadratic equation and solving

Problem: The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 58 / 21 . Find the fraction.

Solution:

Let the numerator of the fraction be x (x be an element of I )

Then its denominator is (2x + 1).

So the fraction is x / (2x + 1))

And the reciprocal would be (2 x + 1) / x

According to the problem:

(X / (2 xs + 1)) + ((2 x + 1) / x) = ( 58 / 21 )

(X ² + (2 x + 1) ²) / (x + (2x + 1)) = ( 58 / 21)

[L. C. D is = ( x + ( 2x + 1 ) ]

21 ( x ² + 4 x ² + 4 x + 1 ) = 58 x ( 2x + 1 )

105 x² + 84 x + 21 = 116 x ² + 58 x

11 x ² - 26 x - 21 = 0

11 x ² - (33 - 7) x - 21 = 0 [using middle term factorization]

11 x ²- 33 x + 7 x - 21 = 0

11 x ( x - 3 ) + 7 ( x - 3 ) = 0

(x - 3) (11 x + 7) = 0

Either ( x - 3 ) = 0 , or ( 11x + 7 ) = 0 [ using zero factor theorem ]

x = 3 ,

From, (11x + 7) = 0

We get, x = - (7 / 11)

But x is an integer , neglect x = - (7 / 11)

Take x = 3

So the required fraction, (x / (2x + 1)) = (3 / (2 * 3 + 1))

= (3 / 7)

Now try the following:

The age of a man is twice the square of the age of his son .Eight years hence the age of the man will be 4 years more than thrice the age of his son. Find their present age?

If you cannot solve this problem, you probably need more practice. A good online tutor would be helpful if you plan to master this subject in a short period. Any reasonably good math tutor should do.