Solving the Quintic

In my previous post about finding the roots of polynomials, I wrote that no general solution exist for polynomials whose degree is higher than 4. However, we still need to be able to find the roots of these polynomials. In this post I want to talk about some of the methods for finding roots, and about their limitations.

Finding the roots of polynomial might sound an abstract goal, but it is actually a very practical one. Look on the following problem for example - Given a line segment of length one AC, divide it in two parts AB and BC in such a way that AC/AB=AB/BC. This is a purely geometrical problem, but its solution is equivalent to solving the following equation:

If we will simplify it a bit, we w! ill get:]

1-x=x^2
x^2+x-1=0

This is a very simple and geometrical example, but the need to solve an equation similar form may arise in different problems in engineering. In this example, it is easy to see that the solution is irrational. This brings us to the first problem with solving polynomials. The answer might be irrational - and if so, even a computer will not be able to solve an equation by trial and error.

Lets look on the following quintic equation (fifth degree polynomial):

This one cannot be solved using a formula. But it is obvious that 1 a root of this polynomial. However, such a polynomial must have five roots (according to the basic theorem of algebra). What can ! we do to find them? A simple way would be to try to divide thi! s polyno mial by x-1. After such division, the degree of the polynomial we need to solve becomes less by one. We will get then that our original polynomial can be written as:

So, we got now a problem which is very easy to solve. All we need to do is to use the formula we know for the forth degree polynomial, and we are done. If we don't remember the formula, we can try to guess another root, lest call it y. and then divide by x-y. For this polynomial, it is again obvious that 1 is a root. So we can divide by x-1 again. We get:

If this is still too complicated, we can divide again.

I wanted the above e! xample to be as simple as possible, so I selected the polynomial (x-1)^5. However, what about more complex examples? It is practical to try to solve some random polynomial in this way? The answer to this is yes. Firstly, we can always divide two polynomials. This result we get from a theorem I don't want to talk about now. Secondly, while it is often impossible to guess the root, this is a rather practical approach. Unless all of the roots are irrational, the chances to guess them are pretty good (especially if you get this question in an exam). The reason for this is that the last monom (the free number) must be multiplication of all the roots, and the second must be minus the sum of all the roots. In the polynomial I solved above, for example the roots are (1,1,1,1,1). There sum is 5 and there multiplication is 1. And in the polynomial this is exactly what we get - in the second place we have (-5x^4) and in the last we have 1.

Lets look on the following polynom! ial:

This particular polynomial cannot be solved by the method specified. The reason for this is very simple - are the roots are complex. For complex roots there is a very interesting property - if, lets say, (2-i) is a root than (2+i) is also a root. This makes the process slightly easier. If you guess one root you get another one as a bonus. Now, I don't know about you, but I have no idea what are the roots of this particular polynomial, and I don't know how to find then.

So why I am saying that the method I presented in which you need to guess a root is practical? It is only practical when you need to solve a polynomial that is "nice and easy". If the polynomial you need to solve cannot be solved by simply guessing, it is better to use the computer to find the solution or at least an approximation to the solution. Unfortunately, this is the only ! thing we can do without a closed formula.

The problem we have with the quintic, and other polynomials is not unique. We have similar problem with integrals. It is often perfectly clear that an integral have a solution, but we have no way to find it. For example:

.

It is proven that this integral cannot be written using elementary functions, but it exists. And there are many others like this one. In modern mathematics there are more than enough things that are proved to be impossible, but we would like them to be possible. And this is (in my opinion) a strong argument for the claim that math is discovered and not invented - we get result which are hardly expected or desired.

In the next post I will write about how simple polynomials were solved in the ancient world, and I will also ! try to find time to write about the formulas for the third and! forth d egree.

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Solve Your Maths Problem with Tutor Vista

Often the case the parents have difficulty in guiding their children at home, especially if the child has difficulty doing math, especially if your child was in K-12. Your child needs help to understand math and algebra concepts, as you know that 90% is to be understood, while the remaining 10% to be memorized. This tutorial now has developed online, allowing your child can follow this tutorial from home via the internet.

In addition, if your child has difficulty with their homework, Tutor vista can also help your child to solve his homework online by using homework help menu. In this tutorial, there are also menus to solve the problems of mathematics, algebra, and trigonometry. Their tutors are always ready and willing to help you anytime on! line 24 hours a day and seven days a week. Tutor vista also provides tutoring for all math problems such as, equation, fractions, simplify and solving equations.
If you still doubt of their service, Tutor Vista also provide demo for free where you can try their services So, What are you waiting for? Just join Tutor Vista and see the result for yourself.

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Word Problems - Hints and Tips to Solving Them

Word problems are one kind of math question that ALWAYS gives students troubles, probably because making a connection between the math and the 'real world' situation isn't necessarily the easiest thing to do. Despite that, there are a few pointers that you can try to which will hopefully lead you to the correct solution.

The first tip is to MAKE A LIST OF ALL OF THE INFORMATION THAT YOU ARE GIVEN.

Example:

Joe wants to save money to buy a new Playstation 3. It costs $400 to buy from the store. He has a job which pays him $15 per hour, and he works 6 hours per day. However, to get to and from his job, he has to take the bus, which costs him $2 each way, and he buys lunch each day he works for $6. He also owes his dad $500 from when he borrowed it to buy a new stereo. Taking all of these costs into account, how many days will it take Joe to save up enough money to buy his new Playstation? How much more money does Joe need to earn if he w! ants to buy it in 5 days?

LIST WHAT YOU KNOW:
PS3 cost = $400
Wages = $15 / hr
Work day = 6 hr / day
Bus fare = 2 x $2 / trip
Lunch = $6 / day
Owes Dad = $500

Next, it is important to IDENTIFY WHAT YOU WANT TO FIND OUT. In this case, we want to find out how many days it will take Joe to save enough to buy his Playstation. Specifically, we want to know how long it will take him to save $400.

The next pointer is to try to CLASSIFY WHAT YOU ARE GIVEN INTO GROUPS. In this question, you can classify things into "Things Related to Making Money" and "Expenses."

Money:

Wages = $15 / hr

Work day = 6 hr / day

Expenses:

Bus fare = 2 x $2 / trip

Lunch = $6 / day

Owes Dad = $500

From here, it might start becoming more apparent what! need to be done. In this case, you would find out how much money Joe makes in a day, and also how much he spends in a day.

Earnings = $15 / hr x 6 hr / day = $90 / day

Expenses = 2 x ($2 / trip) + $6 / day = $10 / day

Overall = 90 - 10 = $80

So, Joe saves $80 per day.

Now, he owes his Dad $500, and wants to buy his Playstation for $400... totalling $900. At $80/day, Joes can pay off his Dad and then be able to afford his Playstation in:

$900 / ($80/day) = 11.25 days.... = 12 days (since he can't stop working at 11.25 days). :)

Of course, now that you have an answer, make sure that you answer the question completely.... there's still an! other part!

If Joe wants to make his purchase, after paying back his dad, in 5 days.....

KNOW:

total cost = $900

time = 5 days

NEED:

savings per day = ?

When you separate what you know from what you don't know, it is very helpful in seeing how to get to the answer. In this case, Joe needs $900 / 5 days = $180. Compared to what he makes right now, he needs to save another $100 per day!

So to summarize, they keys to word problems are to:

IDENTIFY WHAT YOU KNOW (classifying into groups helps here),

DECIDE WHAT YOU NEED TO KNOW,

and MAKE SURE YOU READ AND ANSWER THE ENTIRE QUESTION.

Simple rules, but important none! theless. :)

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Have questions about math problems

Have questions about math problems, get help on this website. You will learn the toughest questions with the easiest way to solve it!!

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Solve x^2 - 10000x - 10000 = 0 without a calculator! A Precalculus Investigation

BACKGROUND/OVERVIEW
No, there's no mistake in the constant term in that equation. Imagine giving this to your Precalculus/Math Analysis/Adv Math students! Actually, 'solving' it with the TI-84 requires some effort using Solver since one needs to make an approximate guess or adjust the lower bound so that the positive root is obtained. Using the graph is no 'walk in the park' either! The TI-89 or Mathematica would have much less difficulty in displaying the exact radical form or a suitable decimal approximation but they may not be within reach. Perhaps an important issue here is that sometimes technology gives us unexpected or even inaccurate results. That's when students need some understanding of theory to recognize the limitations of the technology and adjust accordingly.

Here's the point of all this. The given quadratic is not factorable over the integers, however we can re! place it with a 'nicer' quadratic that is. The roots of the desired quadratic can be shown to be approximately the same as the 'nice' quadratic and we can show that the absolute error is less than two ten-thousanths (and a much much smaller relative or % error)! Does this 'numerical analysis' have any practical value? Why approximate roots when powerful technology can produce exact answers? Do professionals who need to apply mathematics to the solution of 'real' problems ever use such approximation techniques? Could it be that theory actually provides practical application!

THE INVESTIGATION
(1) Show that the roots of the x²-10000x-10001 = 0 are 10001 and -1 by factoring.

! (2) Show that the roots of x2-10000x-10000=0 can be approximated by 10001 and -1 with an error of less than 0.0002.
(a) By direct calculation: Using the quadratic formula and, yes, you may use the calculator!

(b) (Challenging) By comparing, in general,
(*) the roots of x²-bx-(b+1)=0 and
(**) the roots of x²-bx-b=0.
Here we are assuming that b > 0.

(i) First show by factoring that the roots of equation (*) are b+1 and -1.
(ii) Then use the quadratic f! ormula to express the roots of (**) in terms of b.
(iii) Compare the positive roots of these equations by subtracting them and (after algebraic manipulation and simplication), show that the absolute value of the difference is less than 2/(b+1).
Note: For b=10000, this error is therefore less than 0.0002.

(c) Explain intuitively why the roots of the original equation and the 'approximating' equation are virtually the 'same' for 'large' values of b. One possibility here is to consider how the graphs of the associated quadratic functions are related. What do they have in common? How are they different?
Note: Subtle point here for students. Even though the difference of the function values (i.e., y-values) is always 1, this is not true of the difference between their zeros! This may be the essence of the numerical analysis in this i! nvestigation.

EXTE! NSION/PR ACTICE
Ok, now "solve" x² - (googol)x - googol = 0 without a calculator.

RELATED PROBLEM:
Without your calculator show that √(10001) - √(10000) is less than 0.005.
Does this provide us with an effective method of approximating the square root of some large numbers or is it limited and impractical?

For Calculus students: How does this compare to using linearization to approximate the square root?

For more advanced calculus students: Newton's Method? The Binomial Formula (using fractional exponents)? A Taylor Polynomial approximation? All equivalent?

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Using Algebra to Solve Word Problems

Today's lesson notes and the homework - due tomorrow, June 10th (Thursday).

Lesson: Using Algebra to Solve Word Problems

Algebra Word Problems Homework

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An Inexpensive ACT Calculator & a Potpourri of Links

An Inexpensive ACT Calculator & a Potpourri of Links

For the revision to our ACT book, Preparing for the ACT: Mathematics and Science Reasoning, we purchased a cheap scientific calculator from Casio, the fx-115 ES, and I was amazed at the power of this pocket calculator. The calculator sells for about $20 (or less depending o! n where you purchase it), and can:

• work with percents, fractions, and mixed numbers


• simplify radical expressions


• express answers in terms of pi


• do operations with complex numbers


• find permutations and combinations


• perform one-variable statistics (including frequency tables)


• solve one variable equations


• do operations with matrices


• calculate integrals


• do all the usual things that a scientific calculator can do (like radicals, powers, trig functions, etc.)

To top it off, the darn thing runs on solar power! The geek in me can’t help but say, “Wow!” It’s proof that technology can be cheap and powerful (I’m talking to you, Apple.) I called the ACT twice just to confirm that this calculat! or is allowed on the test.

! ACT Tip: If you are planning on taking the ACT test and can’t afford (or don’t want to bother with) a graphing calculator, you can’t go wrong with the fx-115 ES. You are not going to find a TI model with similar capabilities for the same price.

Note/Warning: The ACT is a timed test – you have an average of 1 minute per question. This means two things: (1) Buying a calculator that you are not accustomed to using right before the test is a big mistake. (2) While using an advanced calculator can be helpful in the classroom, you probably won’t have time to use the advanced features on the ACT test. The Casio fx-115 ES’s features that will be useful on the ACT are fractions, mixed numbers, simplifying radical expressions, expressing answers in terms of pi, and operations with complex numb! ers.

To conclude today’s post, here is the promised potpourri of links:

• Engineers beat math PH.D’s in math contest. The contest: The Netflix Prize. Make Netflix’s movie recommendation system more accurate by 10%. The math: statistics.


• We’re all probably going to speak Chinese one day. A group of computer science students from China created one of the most awesome pieces of software I’ve seen in a long time: PhotoSketch (see the video below). It takes a hand-drawn sketch tagged with the name of the object and turns it into a real-world photo. It works in one of those “Why didn’t I think of that?” ways – the software does a web search based on the tags ! and chooses pictures that match the sketch. The best matches a! re then combined together and the user chooses the best looking image. The results are pretty amazing – check out the video below.


• The Making of a Mathlete. PBS is going to air a documentary about the International Math Olympiad. Need I say more? No, really, it actually looks pretty exciting.
  
  

PhotoSketch: Internet Image Montage from tao chen on Vimeo.

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Card Table Playhouse Tutorialette

Okay, I know that some of you have been requesting a card table playhouse tutorial. Since I basically figured it out as I went, I didn't take pictures along the way, and since there's lots of different ways to make the playhouse depending on what you want it to have, I thought we could compromise and I will do a "tutorial-ette". If I were to do it over again I would also do some things differently, so here are my thoughts and tips for the whole thing.

First, here are some pictures of the inside and outside that I did not post earlier.

Infinite processes in the real world

A long time ago, in ancient Greece, one of the philosophers asked a simple yet very important question - is matter infinitely divisible? He of course formulated the question in a much more intuitive way: what will happen if you take a stick and break it in half, than take one of the halves and break it in half again and so on. Thinking about this problem, he concluded that at some point we will not be able to continue breaking the stick. According to him, after a finite amount of time we will reach an indivisible component of matter. He named this indivisible component "atom".
As with any new idea, there were those who believed in it and those who concluded that this idea is wrong. Likely for both sides, there were no means to actually check it so they could argue as much as they wanted.

Even though we are much more advanced today we still don't know the answer to this problem. Ironically we have discovered particles which we named atoms only to find out t! hat they can be split apart as well only a few years later. Although, to be really precise, we need to remember that the problem can be formulated as the "atom" being the basic component of a specific type of mater. In other words, one possible understanding of the problem is that it asks to find a "part" that if divided further looses the recognizable properties of the object we started with. If we formulate the problem in this way, then there are indeed such "atoms" - molecules.

At this point you are probably wondering what is this about and how is it connected to infinity. To understand this lets look on a somewhat famous paradox - the Thomson lamp. Consider a lamp with a toggle switch. Flicking the switch once turns the lamp on. Another flick will turn the lamp off. Now suppose a being able to perform the following task: starting a timer, he turns the lamp on. At the end of one minute, he turns it off. At the end of another half minute, he turns it on again. ! At the end of another quarter of a minute, he turns it off. At! the nex t eighth of a minute, he turns it on again, and he continues thus, flicking the switch each time after waiting exactly one-half the time he waited before flicking it previously. The sum of all these progressively smaller times is exactly two minutes.
So, in the end, is the lamp on or off?

It turns out that there is no clear answer to this problem. While we know the state of the lamp at any time during the process, we cannot tell what is the state at the end. Now lets return to our original problem. Lets suppose for a second that "atoms" don't exist. With this in mind we can take the being from the lamp paradox and instead of it toggling the switch we will make it break sticks in half. Since there are no atoms, the process doesn't end before two minutes pass. But what do we have after two minutes?

In this case it is rather simple to look on the problem mathematically. Lets substitute the stick for the line [0,1]. The whole process can be described t! hen as just a limit of [0,2^(-n)] when n goes to infinity. The limit is a single point, so that would mean that we will get a "particle" with size and mass equaling zero. However, that would suggest that the matter is build from particles with zero mass, and this is a rather bizarre conclusion.
The only possible result we can get from this line of thought is that if such a being actually exists then there are "atoms". However, if there is no such being then we cannot say anything.

While I would like to finish this post with at least a partial solution to the problems I presented, there is no solution as far as I know. There is, however, a funny "solution" to the Thomson lamp paradox. Lets assign numbers to the states of the lamp - 1 and 0. If we do this then the state of the lamp after n steps is: 1-1+1-1+...+(-1)^n.
Therefore, if we take the limit when n goes to infinity, we will get the state of the lamp after two minutes. So lets see what the limit i! s.

A=1-1+1-1+1-...! .
< /div>

1-A=1-1+1-1+1-....=A
2A=1
A=0.5

As you can see, after two minutes the lamp is half on. :)

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Enhancing Student Learning: The Applications of Formative Assessment in Mathematics

In contrast to summative assessment (infrequent, long tests given well after material has been taught), formative assessment enhances learning by motivating students and by providing teachers with more extensive feedback. Consistent, short assessments conducted within a week of learning a new concept are direct manifestations of academic progress to both teachers and students. When students can visually map their progress using the empirical evidence provided by frequent assessments, they are more likely to be motivated and persistent in their studies. Teachers also get a clear picture of the efficacy of their instructional methods. Formative assessment is collaborative, interactive, and positive: three qualities that should characterize any type of education. Translating Formative Assessment into the Language of MathematicsFormative assessment means more than just frequent testing. Teachers must organize and analyze the evidence they gather from ! assessments, using it to develop more informed and successful methods and lesson plans. For math teachers, this means that formative assessments should be developed around goals for student learning such as increased accuracy, deeper understanding of concepts, correct use of vocabulary, logical application of knowledge to new problems, etc. Assessment results should then be analyzed in terms of the goals set by the teacher, and instruction should be modified accordingly. Math-Specific Teaching and Learning StrategiesWritten tests can be effective ways to conduct formative assessment, but oral tests and homework are also important focal points. In any of these cases, it helps to make the assessment more interactive:· Prior to teaching the meaning of new math terminology, have students write down their own definitions for the words. Have them re-write these definitions after instruction. Collect both versions and analyze for basic understanding before building upon th! is knowledge.· When teaching students to solve a new t! ype of p roblem, write out a sample problem on the board along with three or four different answers. Have the students vote on which answer is correct, and then engage students in discussion: which mistakes led to the wrong answers? This strategy helps prevent errors and indicates the depth of students’ understanding of new problem-solving patterns.· Have students write out the steps they take from first viewing a problem to writing down the solution. This can help identify gaps in the problem-solving process so that they can be addressed by further targeted instruction. · Base summative assessments on insights gained from formative assessments. Write tests to analyze documented problem areas and watch for signs of improvement.

Bio: Alexis Bonari is a freelance writer and blog junkie. She is currently a resident blogger at onlinedegrees.org, researching areas of http://www.onlinedegrees.org">onlin! e colleges.

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Free Math Help.com

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Need calculus help?

Do you having trouble on Polynomial Functions, binomial theorem or rational function? I believe some of you having trouble on this precalculus topics, then you need someone that having expertise on precalculus help. If you're a college student you might need calculus help on such topics as integral or differential equation or maybe definite integers.
Based on your grade, if you're having trouble on precalculus or calculus topics, you may try free precalculus help or free calculus help out there. Some paid tutor might provide you with free course so you can decide their quality based on how they help you solving your problem on study. You may also consider to use expert calculus tutor to help you by one on one tutoring about your problem. Having someone to tutor you about precalculus or calculus topics is a good idea but you must make sure that the tutorial is at the pace as your learning speed. Before asking help from expert calculus tutor, make sure the tutor have a good reputation to address your problem correctly and can adapt various matter that his student faced.

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RIPs

UPDATE: Pitchman Billy Mays was found dead today in his home. He was 50. Unbelievable.

I haven't updated in a long time -- no time at all -- but I felt the need to chronicle today/this week as it will become legendary in pop culture history.

Michael Jackson, the "King of Pop," died suddenly today after an apparent heart attack. The artist behind best-selling album of all time died at the age of 50.

Also today, actress Farrah Fawcett lossed her very public battle with cancer. The "Charlie's Angels" actress famous for her pin-up status, Fawcett recently releas! ed a doc umentary depicting her struggles called "Farrah's story." She was 62.

Hollywood lost another legend earlier this week when former Johnny Carson sidekick Ed McMahon submitted to a number of health problems. He was 86.

find the area of the rhombus pictured below.

A factoring trick

I came across the polynomial f(x) = 2x² + 3x - 5 during a calculation I was doing a few days ago. I wanted to factor it. Sure, I could have done it the usual way. But I have a better intuition for factoring numbers than I do for factoring polynomials. So I plug in x = 10; then f(10) = 225.225 factors into 9 times 25. Perhaps this reflects a factorization f(x) = g(x) h(x), where g(10) = 9, h(10) = 25.

Indeed, it does: 2x² + 3x - 5 = (x-1)(2x+5). Of course, this gives a whole family of integer factorizations, plugging in different integers for x.

Of course, this doesn't work in general; consider for example 2x² + 2x + 5, which doesn't factor at all. And when the trick is spelled out explicitly it seems to be irredeemably flawed -- how did I know to take (x-1)(2x+5), say, and not (x-1)(3x-5)? (More importantly, can this be explained without reference to the original polynomial?) One could perhaps point out that, s! ay, 184 = (8)(23), which is just f(9) = g(9) h(9), and so on; from a family of such facts it might be possible to deduce the polynomial factorization, but at that point it's just not worth the trouble. These sorts of tricks, like jokes, rarely stand up to explanation.

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Problem 444: Tangent circles, Secant line, Chords, Angles, Congruence

Geometry Problem
Click the figure below to see the complete problem 444 about Internally Tangent circles, Secant line, Chords, Angles, Congruence.


See also:
Complete Problem 444

Level: High School, SAT Prep, College geometry

tangent of circle

Double tangent line

Find the equation of the line that is Tangent to the Curve
y = x⁴ − 14x³ + 69x² at two points.

Tangent to the Curve

10-1 Simplifying Radicals

simplifying radicals

Nice Try

I feel like I've had the same conversation many times today as we review linear equations. Students have been given an equation in Standard Form and are asked for the x and/or y intercepts. It has gone something like this:

"Mr. Cox, I don't know how to do this."
"Do what?"

"Number 12."
"What's it asking you for?"

"I am supposed to find the x-intercept."
"What's the equation?"

"2x + 3y = 6."
"So what do you know about all x intercepts?"

"They're on the x-axis."
"Alright, then give me an example of an x-intercept."

"5."
"That's a number, give me an x-intercept."

"(4,0)."
"Give me another."

"(-10,0)."
"And one more."

"(7,0)."
"Ok, now what can you tell me about all of those x-intercepts? What do they have in common?"

"y = 0."
"So what do..."

"Oh, that's right, I let y = 0 and solve for x."

It's ! funny how the default is always "I don't get it." Don't let 'em fool ya. They know more than they let on.

Liner equations Question and Answers

Maths - Prime Factors

Methods and tricks to solve questions related to Prime factors.

1). Counting the Number of Factors -- If you factor a number into its prime power factors, then the total number of factors is found by adding one to all the exponents and multiplying those results together.

Example: The total number of factors of 108 are --
108 = (2^2) * (3^3) thus ! we add 1 to the exponents and multiply the results together.

(2+1)*(3+1) = 3*4 = 12.

Hence total number of factors of 108 are 12.

Verifying the above

The factors of 108 are 1, 2, 3, 4, 6, ! 9, 12, 18, 27, 36, 54, and 108 , which are total 12 in number.!


2). Factoring Factorials - First write the number you are factoring as a product of two or more numbers. For example, suppose we want to factor the number 30.

We know that 30 = 5 x 6,

so we write for the first step of our factor tree:

30
/ \
5 6

Next we factor the factors, if possible. In other words, for each number in the product from the first step, we try to write it d! own as a product of even smaller numbers.

For instance, in our example, we would try to factor both 5 and 6. As we noted above, 5 is prime, so we can't factor it further.

We are done with that branch of the factor tree.

However, 6 is not prime. 6 = 2 x 3, so we can extend our factor tree as follows:

30
/ \
5 6

Now further

6
/ \
2 3

Now, we continue this process until all of the branches of the tree end in prime numbers.

In our example, we are done after two steps, since 5, 2, and 3 are all prime numbers.

The factors of the number are the numbers at the end of the different branches on the factor tree. To figure out what power each prime factor is raised to, count the number of times the prime factor appears in the factor tree.

In our example, each of the factors appears only once, so the prime factorization of 30 is: 30 = 2 x 3 x 5.

prime factorization

Graphing - Equation of the Line

If you are given a line on a graph (or enough information to construct a line), you will likely also be asked to find the equation of the line. The equation of the line is unique to each line; that is, every line has a different equation. With it, you can readily tell the slope of the line, and you can calculate what the x-value is for any y-value on the line (and vice versa). It is very handy!

The most common and basic form of the equation of the line is:

y = mx + b

where m is the slope and b is the y-intercept (where the line crosses through the y-axis).

Another way to write it, which is more general is called the POINT-SLOPE FORMULA:

(y-y1) = m(x-x1)

where m is the slope, and x1 and y1 are the coordinates of a known point on the graph. (If you pay close attention, you can see that this way of writing it is really the same as t! he slope formula, but rearranged!) The version basically says 'give any point at all, and a slope, and you have enough information to draw the line."

Let's use the graph from the slope lesson to practice, using points (2,1) and (7,7). To get our final answer, we are going to have to do a couple of steps first. Let's use the y=mx+b equation. The steps we are going to do are:
1) Find the slope
2) Find the y-intercept
3) Write the equation of the line

Finding the slope is easy now, if you read the posting on slopes (I always leave numbers as fractions, instead of changing to decimals):

m = (y2-y1)/(x2-x1)
= (7-1)/(7-2)
= 6/5

The y-intercept is easy now... just plug numbers into y=mx+b, and solve for b. So, b=y-mx. We have our slope now, and for y and x, we ju! st substitute in the coordinates of a single point (x and y MUST be from the same point!)

b=y-mx
= 1-(6/5) x 2
= 1-12/5
= (-7/5) (you have to do some fraction math!)

So now we can write the equation of our line!

y = mx+b
y = (6/5)x - (7/5).... (if we leave it like this, we can read the values for slope and y-int right from the equation!)
y = (6x - 7)/5

If we work through using the other formula... (y-y1) = m(x-x1)... we get the same answer, and we don't have to explicitly solve for b! First, solve for slope. Second, substitute in values for slope and x1,y1, and then rearrange!

slope = 6/5 (same thing we did above)

so (y-y1) = m(x-x1)..... (sub in slope, and point (2,1) for (x1,y1))
(y-1) = (6/5)(x-2)...
y=(6/5)(x-2) + 1...
y = (6/5)x - (6/5)2 + 1...
y=(6/5)x -12/5 + 1...
y=(6/5)x -(7/5)... (same slope ! and y-int)
y=(6x-7)/5

Same answer! That is the equation of the line. Now if you put in any value for x, you can say exactly what the value for y would be... if you wanted to know what y is when x is 1000, you can do that now! You will see that it will always be on the same straight line.

If we work through and do the same thing for the red line on the graph, with points (-6,2) and (-4,-5), we can get the equation for that line too! Let's try it this time using just the one formula.

(y-y1) = m(x-x1)
((-5)-2) = m((-4)-(-6))
(-7) = m(2)
m=(-7/2)

Now put that back into the same formula along with a point, and rearrange:

(y-y1) = m(x-x1)
(y-2) = (-7/2)(x-(-6))
y-2 = (-7/2)(x+6)
y= (-7/2)x + (-7/2)(6) + 2
y = (-7/2)x -21 + 2
y = (-7/2)x -19

(see slope (-7/2) and y-int (-19)... look at the graph, and you will see that makes sense! Negative slope, very low y-int!)

That's all there is to it! Just remember that there a! re a few steps to follow, depending on the formula you are using... basically, remember to always find the slope first, then the y-intercept, and plug directly into y=mx+b... or use the point-slope formula to find the slope using 2 points, and then resubstitute it back in with a single point and rearrange. It's not that complicated once you practice and understand what you are doing! Either method is going to give you the same answer, so pick your favorite and stick with it!

find equation of a line

Guitar Lesson- Circle of Fourths Chord Progressions

 

Chord progressions using the circle of fourths are popular in all musical styles. Starting from the first chord in the progression, each subsequent chord will be a fourth higher in the key. The circle of fourths utilizes every chord in the key, playing through them one by one.

The following progression contains chords from C major. This progression is actually in the key A minor. C major and A minor both contain the same exact notes, making them relative keys. The key of A minor is considered the relative minor of C major. Conversely, C major is considered the relative major of A minor. Using this knowledge, you can play solos in relative keys, using the notes of the Am pentatonic scales over a progression in C major.

Notice that A is the sixth step of the C major scale. In any major key, the relative minor scale starts on the sixth step. This is also called a mode of the scale. There is a different mode for each step of the scale. 

Click on the image to enlarge.

Here is the circle of fourths chord progression in Am, presented as a rhythm in tablature. Practicing this progression will help you become fluent in changing from chord to chord using the full form chords from the previous lesson. Once you’ve got this rhythm down, try to vary it and come up with your own strumming patterns. You can also record this rhythm and play solos over it using all of the C major sc! ales, since C major and A minor are relative keys.

Click on the image to enlarge.

©2009 Fred Russell Publishing, All Rights Reserved. This article can not be used without permission from the Author. To Contact the Author email curt@RockHouseMethod.com

Guitar Lesson- Circle of Fourths Chord Progressions

chord of a circle

Guitar Lesson- Circle of Fourths Chord Progressions

 

Chord progressions using the circle of fourths are popular in all musical styles. Starting from the first chord in the progression, each subsequent chord will be a fourth higher in the key. The circle of fourths utilizes every chord in the key, playing through them one by one.

The following progression contains chords from C major. This progression is actually in the key A minor. C major and A minor both contain the same exact notes, making them relative keys. The key of A minor is considered the relative minor of C major. Conversely, C major is considered the relative major of A minor. Using this knowledge, you can play solos in relative keys, using the notes of the Am pentatonic scales over a progression in C major.

Notice that A is the sixth step of the C major scale. In any major key, the relative minor scale starts on the sixth step. This is also called a mode of the scale. There is a different mode for each step of the scale. 

Click on the image to enlarge.

Here is the circle of fourths chord progression in Am, presented as a rhythm in tablature. Practicing this progression will help you become fluent in changing from chord to chord using the full form chords from the previous lesson. Once you’ve got this rhythm down, try to vary it and come up with your own strumming patterns. You can also record this rhythm and play solos over it using all of the C major sc! ales, since C major and A minor are relative keys.

Click on the image to enlarge.

©2009 Fred Russell Publishing, All Rights Reserved. This article can not be used without permission from the Author. To Contact the Author email curt@RockHouseMethod.com

Guitar Lesson- Circle of Fourths Chord Progressions

chord circle

Geometry Problems 71-80: Cyclic Quadrilateral, Intersecting Circles, Area, Inradius

Geometry Problem
Click the figure below to see problems 71-80 about Cyclic Quadrilateral, Intersecting Circles, Area, Inradius, Incircle, Circular Sector, Parallel, Chord, Secant.


See also:
Problems 71-80

Level: High School, SAT Prep, College geometry

area of a quadrilateral problems

Free algebra worksheets

Usually algebra textbooks provide lots of problems to practice the algebraic concepts and techniques, but some of you may still benefit from resources for free (or mostly so) printable algebra worksheets. Please see the list below, which I've originally compiled for my HomeschoolMath.net site.

Algebra worksheets

Worksheet Builder
Great and free worksheet maker software with nearly 7,000 built-in algebra and geometry questions.
www.jmap.org/JMAP_WORKSHEET_BUILDER_INSTALLATION_FILES.htm

Free Algebra Worksheets from KUTA Software
Free worksheets (PDF) for equations, exponents, inequalities, polynomials, radical & rational expressions and more.
www.kutasoftware.com/free.html

AlgebraHelp.com worksheets
Interactive worksheets that are checked online for most algebra 1 topics.
www.algebrahelp.com/worksheets/

Math.Com algebra worksheets generator
Generate worksheets for: linear equations, systems of equations, and quadratic equations.
www.math.com/students/worksheet/algebra_sp.htm

LessonCorner worksheets
These free worksheets include a few topics such as calculations with polynomials, factoring, and graphing lines.
www.lessoncorner.com/worksheets/

Algebra Fun Sheets
Worksheets that integrate algebra skills with fun activities including sudoku, word finds, riddles, color patterns, crosswords, games, matching cards, etc. A subscription is required.
www.algebrafunsheets.com

About.com Algebra Worksheets
An assorted collection of free algebra worksheets and answers. These pages are not very well organized, but they have lots of worksheets.
math.about.com/od/algebraworksheets/Algebra_Worksheets.htm

Algebra Worksheets from MathWorksheetCenter
Lots of worksheets for over 100 algebra topics. A few are free; most are accessible only by one-year a subscription.
www.mathworksheetscenter.com/mathskills/algebra/

A few fun algebra worksheets
These are for graphing linear equations and linear inequalities.

Online Math Work
Free multiple-choice worksheets for pre-algebra and algebra 1 topics. You can do them online, or copy to a word processor to print.
www.mathonlinework.com

Lastly... my own algebra worksheet collections, which aren't free but there are many free samples:

Math Mammoth Algebra 1 Worksheets Collection
A two-part collection (A and B) of 137 quality algebra worksheets covering all the topics in a typical algebra 1 curriculum. These worksheets are hand-crafted and contain lots of word problems and other variable problems. Free samples available. $11.50.
www.mathmammoth.com/worksheets/algebra_1.php

free pre algebra answers

A free download of a digital Algebra 1 book

Kinetic Books Algebra 1 looks really interesting! It is not really just a book, but software, or a digital interactive textbook.

It contains text, interactive problems and activities, and a scoring system all on the computer. Students can get step-by-step assistance in the form of audio hints and one-click access to relevant examples.

See a demo here. But the best is that the company Kinetic Books is even offering a free download of the product till September 30!

That really sounds fantastic, so if you have algebra 1 student(s), don't fail to take advantage of this tremendous offer.

free algebra homework help

Pre-Algebra and My Dear Aunt Sally

Will you Please Excuse My Dear Aunt Sally? It seems she had indigestion and burped. Can you find where she is? This is how I had fun publishing my Unit on Pre-Algebra! This unit includes in-depth lessons on:

1. Order of Operations
2. Order of Operations with Exponents
Order of Operations with Integers
3. Writing AlgebraicExpressions
4. Writing AlgebraicEquations
5. Practice Exercises
6. Challenge Exercises

So can you find my Dear Aunt Sally? And will you excuse her?

online pre algebra help

Factoring the time

I stumbled upon this comic that you might enjoy... factoring the time (from xkcd.com).

free online math help

Free algebra worksheets

Usually algebra textbooks provide lots of problems to practice the algebraic concepts and techniques, but some of you may still benefit from resources for free (or mostly so) printable algebra worksheets. Please see the list below, which I've originally compiled for my HomeschoolMath.net site.

Algebra worksheets

Worksheet Builder
Great and free worksheet maker software with nearly 7,000 built-in algebra and geometry questions.
www.jmap.org/JMAP_WORKSHEET_BUILDER_INSTALLATION_FILES.htm

Free Algebra Worksheets from KUTA Software
Free worksheets (PDF) for equations, exponents, inequalities, polynomials, radical & rational expressions and more.
www.kutasoftware.com/free.html

AlgebraHelp.com worksheets
Interactive worksheets that are checked online for most algebra 1 topics.
www.algebrahelp.com/worksheets/

Math.Com algebra worksheets generator
Generate worksheets for: linear equations, systems of equations, and quadratic equations.
www.math.com/students/worksheet/algebra_sp.htm

LessonCorner worksheets
These free worksheets include a few topics such as calculations with polynomials, factoring, and graphing lines.
www.lessoncorner.com/worksheets/

Algebra Fun Sheets
Worksheets that integrate algebra skills with fun activities including sudoku, word finds, riddles, color patterns, crosswords, games, matching cards, etc. A subscription is required.
www.algebrafunsheets.com

About.com Algebra Worksheets
An assorted collection of free algebra worksheets and answers. These pages are not very well organized, but they have lots of worksheets.
math.about.com/od/algebraworksheets/Algebra_Worksheets.htm

Algebra Worksheets from MathWorksheetCenter
Lots of worksheets for over 100 algebra topics. A few are free; most are accessible only by one-year a subscription.
www.mathworksheetscenter.com/mathskills/algebra/

A few fun algebra worksheets
These are for graphing linear equations and linear inequalities.

Online Math Work
Free multiple-choice worksheets for pre-algebra and algebra 1 topics. You can do them online, or copy to a word processor to print.
www.mathonlinework.com

Lastly... my own algebra worksheet collections, which aren't free but there are many free samples:

Math Mammoth Algebra 1 Worksheets Collection
A two-part collection (A and B) of 137 quality algebra worksheets covering all the topics in a typical algebra 1 curriculum. These worksheets are hand-crafted and contain lots of word problems and other variable problems. Free samples available. $11.50.
www.mathmammoth.com/worksheets/algebra_1.php

algebra 1 help free