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.

Algebra DVDs - An Excellent Learning Tool to Teach Your Kids Algebra

Algebra is the line of demarcation between basic and higher math. Many children panic upon seeing letters and numbers in a mathematical equation. They often run to their parents for help. Parents there is a resource specifically designed to help you teach your kids algebra, Algebra DVDs.

Algebra DVDs can be a parent's best friend. They provide simple step by step instructions on the basics of algebra along with simple exercises allowing parent and child to test their mastery of each topic. The taped lessons use cartoons, songs, and real world examples to reinforce the concepts being taught.

One of the greatest gifts a parent can give a child is to show them your willingness to learn even though you are an adult. This helps children to understand that not knowing something is not bad as long as you are willing to admit you don't know and seek out help.

A parent and child working their way through this teaching tool together can help to strengthen their bond while increasing their mastery of the subject matter.

Mastering mathematics requires a solid understanding of Algebra. Scientist, engineers, business people and industrialists use algebra to solve problems every day. Understanding Algebra gives children more career choices.

Parents can use the guided lessons to help their children understand basic terms like variable, equation, and factors and follow them through the more complex concepts.

One of the great strengths of the Algebra DVDs is the ability to replay them endlessly until you understand a concept. Standard classroom instruction does not allow you to do this. So buy Algebra DVDs, teach your child algebra and watch their confidence grow.

Factoring Trinomials (Quadratics) - Lucid Explanation of the Method With Examples

Consider the product of the two linear expressions (y+a) and (y+b).

(y+a)(y+b) = y(y+b) + a(y+b) = y^2 + by + ay + ab = y^2 + y(a+b) + ab

We can write it as

y^2 + y(a+b) + ab = (y+a)(y+b) .......(i)

Similarly, Consider the product of the two linear expressions (ay+b) and (cy+d).

(ay+b)(cy+d) = ay(cy+d) + b(cy+d) = acy^2 + ady + bcy + bd = acy^2 + y(ad+bc) + bd

We can write it as

acy^2 + y(ad+bc) + bd = (ay+b)(cy+d) .......(ii)

Equation (i) is Simple Quadratic Polynomial expressed as Product of Two linear Factors and Equation (ii) is General Quadratic Polynomial expressed as Product of Two linear Factors

Observing the two Formulas, leads us to the method of Factorization of Quadratic Expressions.

In Equation (i),

the product of coefficient of y^2 and the constant term = ab

and the coefficient of y = a+b = sum of the factors of ab

Similarly, In Equation (ii),

the product of coefficient of y^2 and the constant term = (ac)(bd) = (ad)(bc)

and the coefficient of y = (ad+bc) = sum of the factors of acbd

So, if we can resolve the product of y^2 and the constant term into product of two factors in such a way that their sum is equal to the coefficient of y, then we can factorize the quadratic expression.

We discuss the steps involved in the method and apply it to solve a number of problems.

Method of Factoring Trinomials (Quadratics) :



Step 1 :

Multiply the coefficient of y^2 by the constant term.

Step 2 :

Resolve this product into two factors such that their sum is the coefficient of y

Step 3 :

Rewrite the y term as the sum of two terms with these factors as coefficients.

Step 4 :

Then take the common factor in the first two terms and the last two terms.

Step 5 :

Then take the common factor from the two terms thus formed.

What you get in step 5 is the product of the required two factors.

The method will be clear by the following Solved Examples.

The examples are so chosen that all the models are covered.

Example 1 :

Factorize 9y^2 + 26y + 16

Solution :

Let P = 9y^2 + 26y + 16

Now, follow the five steps listed above.

Step 1:

(Coefficient of y^2) x (constant term) = 9 x 16 = 144

Step 2:

We have to express 144 as two factors whose sum = coefficient of x = 26;

144 = 2 x 72 = 2 x 2 x 36 = 2 x 2 x 2 x 18 = 8 x 18; (8 + 18 = 26)

Step 3:

P = 9y^2 + 26y + 16 = 9y^2 + 8y + 18y + 16

Step 4:

P = y(9y + 8) + 2(9y + 8)

Step 5:

P = (9y + 8)(y + 2)

Thus, 9y^2 + 26y + 16 = (9y + 8)(y + 2) Ans.

Example 2 :

Factorize y^2 + 7y - 78

Solution :

Let P = y^2 + 7y - 78

Now, follow the five steps listed above.

Step 1:

(Coefficient of y^2) x (constant term) = 1 x -78 = -78

Step 2:

We have to express -78 as two factors whose sum = coefficient of y = 7 ;

-78 = -2 x 39 = -2 x 3 x 13 = -6 x 13; (-6 + 13 = 7)

Step 3:

P = y^2 + 7y - 78 = y^2 - 6y + 13y - 78

Step 4:

P = y(y - 6) + 13(y - 6)

Step 5:

P = (y - 6)(y + 13)

Thus, y^2 + 7y - 78 = (y - 6)(y + 13) Ans.

Example 3 :

Factorize 4y^2 - 5y + 1

Solution :

Let P = 4y^2 - 5y + 1

Now, follow the five steps listed above.

Step 1:

(Coefficient of y^2) x (constant term) = 4 x 1 = 4

Step 2:

We have to express 4 as two factors whose sum = coefficient of y = -5 ;

4 = 4 x 1 = -4 x -1; [(-4) + (-1) = -5]

Step 3:

P = 4y^2 - 5y + 1 = 4y^2 - 4y - y + 1

Step 4:

P = 4y(y - 1) - 1(y - 1)

Step 5:

P = (y - 1)(4y - 1)

Thus, 4y^2 - 5y + 1 = (y - 1)(4y - 1) Ans.

Example 4 :

Factorize 3y^2 - 17y - 20

Solution :

Let P = 3y^2 - 17y - 20

Now, follow the five steps listed above.

Step 1:

Coefficient of y^2 x constant term = 3 x -20 = -60

Step 2:

We have to express -60 as two factors whose sum = coefficient of x = -17 ;

-60 = -20 x 3; (-20 + 3 = -17)

Step 3:

P = 3y^2 - 17y - 20 = 3y^2 - 20y + 3y - 20

Step 4:

P = y(3y - 20) + 1(3y - 20)

Step 5:

P = (3y - 20)(y + 1)

Thus, 3y^2 - 17y - 20 = (3y - 20)(y + 1) Ans.

Example 5 :



Factorize 2 - 5y - 18y^2

Solution :

Let P = 2 - 5y - 18y^2 = -18y^2 - 5y + 2

Now, follow the five steps listed above.

Step 1:

(Coefficient of y^2) x (constant term) = -18 x 2 = -36

Step 2:

We have to express -36 as two factors whose sum = coefficient of y = -5 ;

-36 = -2 x 18 = -2 x 2 x 9 = 4 x -9; [4 + (-9) = -5]

Step 3:

P = -18y^2 - 5y + 2 = -18y^2 + 4y - 9y + 2

Step 4:

P = 2y(-9y + 2) + 1(-9y + 2)

Step 5:

P = (-9y + 2)(2y + 1)

Thus, 2 - 5y - 18y^2 = (-9y + 2)(2y + 1) Ans.

Example 6 :

Factorize (y^2 + y)^2 -18(y^2 + y) + 72

Solution :

Let P = (y^2 + y)^2 -18(y^2 + y) + 72

Put (y^2 + y) = t; Then P = t^2 -18t + 72

Now, follow the five steps listed above.

Step 1:

(Coefficient of t^2) x (constant term) = 1 x 72 = 72

Step 2:

We have to express 72 as two factors whose sum = coefficient of t = -18 ;

72 = 12 x 6 = -12 x -6; [(-12) + (-6) = -18]

Step 3:

P = t^2 -18t + 72 = t^2 - 12t - 6t + 72

Step 4:

P = t(t - 12) - 6(t - 12)

Step 5:

P = (t - 12)(t - 6)

But t = (y^2 + y);

So, P = (t - 12)(t - 6) = (y^2 + y - 12)(y^2 + y - 6)

In each of these two brackets, there is a Quadratic Polynomial which can be factorised using the five steps above.

y^2 + y - 12 = y^2 + 4y - 3y - 12 = y(y + 4) - 3(y + 4) = (y + 4)(y - 3)

y^2 + y - 6 = y^2 + 3y - 2y - 6 = y(y + 3) - 2(y + 3) = (y + 3)(y - 2)

See how these two Quadratic Polynomials are factorised with the knowledge of the 5 steps.

You might have mastered the 5 steps of factorisation by this time, to write directly like this.

Thus,

P = (y^2 + y)^2 -18(y^2 + y) + 72

= (y^2 + y - 12)(y^2 + y - 6)

= (y + 4)(y - 3)(y + 3)(y - 2) Ans.

Algebra: How To Start and Why You Really Need to Practice a Lot

The most importand factor about algebra and mathematics is that you really need to practise in order to understand the concepts. There are a lot of algebra books, tutorials, interactive lessons, online tutors and courses that can help you a lot, but without some homework and practise you are bound to fail.

You cannot remember the thousands of algebra concepts and equations without truly understand what each equation really means. What is the reason behind every algebra concept.

A key factor is to take small steps. For example learn one algebra equation, practise it for a couple of days, then move on to the next. Move on only if you feel comfortable with what you've learned. Every time you move on to the next algebra concept increase the difficulty of the equations or the exercises you're trying to solve.

Follow this procedure and soon you'll notice that your brain will be able to absorb mathematics and algebra quicker and easier. You will start to like it.

Another thing is that algebra requires you to apply specific rules. You will find these basic rules in a lot of equations and algebra concepts. The more you practise the basic algebra rules the better you'll become at learning different and more difficult algebra concepts.

A lot of people learning algebra miss out on the basic rules. In the end they mess up everything. Remember, rules are everywhere. Our society is based on specific rules. Same goes for algebra. Treat algebra rules like a game and you'll like it.

You may end up wondering.. Why should I learn math and algebra? That's a good question. You must ask yourself this question before you go on. Algebra will help you solve problems, make decisions, apply strategies and a lot more. If you want your skills to worth gold then you have to learn math and especially algebra. If you want your mental powers to reach top level then you have to learn algebra.

If you like mathematics then you know that in order to move to advanced mathematics you have to know the basics of algebra. Algebra is everywhere in your everyday life. This is a great advantage when trying to learn algebra. Why? Because you can take the a real situation and turn it into an algebra concept or equation.

Always ask yourself.. why? Here is a simplified example. You walk on the street and you see a girl carrying a birdcage. Inside the birdcage there are 4 little birds. Suddenly the cage's door opens and one of the birds escapes. You can turn this incidence into a simple algebra equation:

3 - 1 = 2

See what I mean? Now that's the first step! After that it will be easy to turn live complicated situations into algebra concepts. Your brain will soon become familiar with the idea. And soon you will be able to turn your entire life into algebra!

Algebra Students: Vocabulary Is Important In Math Class, Too! Ask If You Are Confused About Meanings

I recently read a blog from a math teacher whose main area of concern is elementary math and literacy. I had to laugh at the example he used--why do we use the word "borrow" in subtraction when we don't intend to give it back? This is a point well-taken. We tend, in mathematics, to use terms from everyday speech, but in mathematics, these terms often have very different meanings. This can be and often is a source of confusion for many Algebra students; and, unfortunately, students often don't even realize they have confused the meanings of a term, and the teacher doesn't catch it is until too late. By "too late" I mean that the mistake has been practiced and has become ingrained in your brain as a fact. These mistakes are difficult to fix. It is better to avoid these than try to fix them.

One term from your elementary days that still causes students difficulty is the term "value"--especially with respect to fractions. If I asked you, "Is 3/4 > 1/2" what you say? You and almost everyone else would say "Yes." In reality, the answer is "not necessarily." The problem here is that fraction symbols do not actually have a VALUE until you know the "of what." Is 3/4 of an inch greater than 1/2 of a foot? Of course not, you say? Why not? You just told me 3/4 > 1/2.

Now that you are thinking a little bit more about it, you realize that fractions can only be compared IF they are fractions of the SAME THING. So why have math books had homework sections with instructions to "compare the fractions?" Because someone assumed you knew that fractions can only be compared if they are of the same thing, and they left out an important part of the instructions. "Compare these fractions on a number line" or "Assume these fractions are of they same thing" would be appropriate directions. On a number line, 3/4 and 1/2 represent parts of the same size unit. We math teachers tend to assume that every student is picturing the same thing or understanding a definition the same way that we intend; but you and I both know this isn't always true.

In Algebra, there are two huge examples of extremely important concepts that students often get confused with their everyday meanings--or at the least, cannot really explain what the math meaning actually is. These two concepts are: "solve" as in solve an equation and "factor" as in factor this expression.

You know what it means to solve a puzzle or to solve a problem you are having making free throws in basketball; but what does it mean to solve an equation? To find the answer, you say? How do you know when you have an answer? It works? What does that mean? Very few Algebra students can actually say in words--with any real understanding--that to solve an equation means to find values for the variables that make the equation TRUE.

You know that washing hands is an important "factor" in slowing the spread of disease; but how does this apply to factoring an expression like a^2 - ab? The mathematics meaning of "factor" is totally different from the everyday meaning. In Algebra, to factor means to "re-write as multiplication." What? Well, a^2 - ab in factored form is a(a-b) since when you multiply a and (a - b) you get a^2 - ab.

All of mathematics--not just Algebra--is full of these terms with different meanings in the everyday world than in the mathematical world. For your own success, you must always memorize math definitions immediately, practice these definitions, and even discuss with yourself and with your teacher the differences in meanings. It is OK--in fact, important--to know that a term has several different meanings. It is equally important that you understand each of the meanings and know when to use which meaning.

If you get confused, or are ever in doubt, ASK YOUR TEACHER! It is the teacher's responsibility to teach you. We math teachers are not perfect human beings, even if we like to think we are. We often ASSUME more than we should. DO NOT BE AFRAID TO ASK QUESTIONS. That is your responsibility. Ours is to answer your questions.

Algebra Made Easier With Algebra 2 Homework Help!

Algebra 2 is a step beyond Algebra 1. Before commencing this stage in the educational process, students need to be thoroughly grounded with the foundations. Algebra 2 homework help can answer some of the basic questions about Algebra. But is it advisable that students have tried to do their personal Algebra homework before consulting with this source. In Algebra 2, students are introduced to logarithms and exponents, graphic functions, ways of solving inequalities and equations with complex numbers. The course structure also includes polynomial arithmetic, rational expressions, radicals and complex numbers, quadratic system and cone sections. As new terms, these words might sound a little complicated. However, Algebra 2 homework help becomes a personal guide to make complex mathematics easier for students.

Help for competitive examinations

Algebra 2 homework help is a very effective platform when preparing for competitive examinations or a university entrance examination. A foundation in this category of mathematics is a must when a student is applying for the General Educational Development examination. The GED exam earns the taker the equivalent of a high school diploma, which is necessary for students who cannot complete their high school courses. The university entrance exams are the SAT and ACT. Students need not bother to seek out a personal teacher. All the queries are easily resolved with the resources found at the Algebra 2 homework help site.

Benefits of the Algebra 2 Homework help platform

This platform's motto is to make mathematics easier for students. By availing themselves of this service, students can get rid of the complexities of the subject. Algebra 2 homework help provides tips for the effective study of mathematics. The students have to possess sharp memory skills in order to excel in this section of math. The help book provides certain very interesting ways of making memory skills sharper to assist the learning process. This is also helpful for students who do not remember the elementary lessons of Algebra. It is not possible to provide a detailed version here, but a basic reference is provided. This reference helps students brush up their memories to remind them of the lessons that are learned in Algebra 1.

Algebra 2 homework help trains students to first understand the problem and then locate the most efficient way to solve it. For instance, when looking to solve an equation, the first step is to observe the number of terms in the given equation. The next step, then, is to decide which type of factoring to opt for to solve the equation. Algebra 2 homework help also works on a student's critical thinking capability. This process helps in case students want to take up mathematics for further higher studies. It helps students gather a sound knowledge required in order to deal with complex mathematical problems at higher levels. It is as easy and friendly as troubleshooting.

Career prospects made attractive with Algebra 2

People who excel in Algebra have great career prospects. They might not be aware of this, but they can get really high salaried jobs. They can even join any educational institute and spread their knowledge.

Why Study Algebra? Completing the Square It's Not As Hard As You Think

One of the most useful techniques in algebra is that of completing the square. The name is appropriate as the geometric interpretation encompasses the formation of a square from a rectangle by the addition of an appropriate quantity. Geometry aside, this technique has many applications, not only in algebra, but also in more advanced realms such as integration, which is a key component of integral calculus. Here we will see that this technique can be had rather inexpensively.

Completing the square involves taking a non-perfect square trinomial and converting it into a perfect square. Actually, this technique is performed when you have a quadratic equation set to zero, as in x^2 + 10x - 5 = 0. If you recall, a perfect square trinomial is one in which the middle coefficient is equal to twice the product of the square roots of both the leading coefficient and the constant term. What a mouthful! Let's look at a specific example. Take the quadratic trinomial x^2 + 10x + 25. The leading coefficient is 1, the number (which is understood) in front of the x^2 term. The middle coefficient is 10, and the constant term is 25. The square root of 1 is naturally 1; the square root of 25 is 5; 2*1*5 is 10, which is the middle coefficient. Thus x^2 + 10x + 25 qualifies as a perfect square trinomial.

So what is so special about these trinomials? Well for one, they can always be factored into the form (x +/- c)^2. In other words, we can always factor them as (x + c)^2 or (x - c)^2, where c is a constant and the "+" or "-" is dictated by the sign of the middle coefficient. Once factored, we can easily solve any quadratic equation by performing the simple operation of taking the square root and adding or subtracting the constant c. To see this, let us look at a specific example.

Suppose we wish to solve the quadratic equation x^2 + 8x - 10 = 0. You cannot solve this by factoring. You can of course go directly to the quadratic formula, but an even quicker way is to complete the square, and this is how we shall do it. Isolate the x-terms, namely x^2 and 8x, on one side of the equation and bring the constant term to the other. Remember that when we move the -10 over we get +10. Thus we have x^2 + 8x = 10. Now begin the process of converting x^2 + 8x into a perfect square. We take half of 8, which is 4 and square it to get 16. We add this quantity to both sides of the equation to get x^2 + 8x + 16 = 10 + 16 = 26. Now if you check the conditions which make a trinomial perfect, you will see that x^2 + 8x + 16 fits the bill. That is 2*4*1 = 8.

Since the trinomial is now perfect, we can factor it into (x + 4)^2, that is we take the x term, half of 8, and the "+" sign, since the middle term is positive. We write (x + 4)^2 = 26. To solve this equation, we simply take the square root of both sides, remembering to take the "+" and "-" part. (Remember: when we take a square root in an equation, we always consider both the positive and negative values). Thus we have (x + 4) = +/- the square root of 26. (Since I cannot use the square root symbol in this article, I will write 26^.5 as the square root of 26; actually this is true since the square root is the one-half power.) To finish this off, we subtract the 4 from both sides to solve for x, and we get x = -4 +/- (26)^.5, that isx = -4 + (26)^.5 or x = -4 - (26)^.5. Since (26)^.5 is equal to a little more than 5, about 5.1, we have that x is equal to about 1.1 or -9.1.

With this technique, you can now solve any quadratic, regardless of whether it is factorable or not, without resorting to the quadratic formula. To sum up, all you need do is the following (As you read these steps, refer back to the example just done):

1) Isolate the x terms on one side of the equation and the constant term on the other;

2) Take half the middle coefficient, square it and add it to both sides of the equation;

3) Factor the trinomial using (x +/- c)^2, where c is equal to half the middle term, and the sign is taken according to the sign of the middle coefficient; and

4) Take the square root of both sides, remembering to consider the +/- cases, and add or subtract c to both sides.

With the ammunition given above, you are now expert at completing the square and solving any quadratic equation. Isn't life grand!