3+1 LCC/Ferris State Education Program

For those of you that live in the Lansing area..... There is a new 3+1 LCC/Ferris State Program for Education. The program also can be with an emphasis in Mathematics. That is what I've decided to pursue. The 3+1 program works out to be 3 years at Lansing Community College with their rates and the last year at Ferris State University. This way you can save a ton of $$$$ and I hear it is a good program. I only have to go up to Calculus 2 which I did 6 years ago, but I also have a 300 level Geometry class that I am going to take also and am excited about. So for those of you who live in the Lansing, MI area and want to pursue an Education degree whether it be in Math or not, don't hesitate to look up this neat program of which I believe will suit me and others well.

Finding the domain of a square root function

It has been a long time since I've posted. I apologize for this inconvenience. I will also attempt to answer questions. I enrolled at LCC. It has been 5 years since I've taken a Calculus course, so I've gone back to college Algebra. Since I've gone back to College Algebra, I've reviewed and learned some new things, here is an interesting problem: Solve for x.

f(x)=sqrt x^4-26x^2+25

The first thing you would do is: FACTOR the polynomial into linear factors. From the leading-term test we know that f(x) has at most 4 asymptotes since it is of the fourth degree. Also since it is of the fourth degree we know the beginning and end behaviour of the graph. We know it is concave up since the leading term is positive and that it is a parabola shape since the degree is even. But now we have to factor the polynomial into linear factors. The factor theorem states that:

For a Polynomial of f(x) if f(c)=0 , then x-c is a factor of f(x). Simply put if f(3) =0 then (x-3) is a factor for an example. We can factor this by stating the possible rational zeros of the function. The possible rational zeros of the function are P/Q. Which is constant term over coefficient of leading term. Therefore the possible rational zeros are: +1,-1,+5,-5,+25,-25. -1 happens to be a zero, therefore we find the 3rd degree equivalent polynomial with synthetic division

-1

1 0 -26 0 25

-1 1 25 -25

1 -1 -25 25 0

Therefore these are our coefficients to our 3rd degree polynomial. x^3-x^2-25x+25=0

-5 is a zero of this equivalent polynomial. Therefore use synthetic division again to get a second degree polynomial.

-5

1 -1 -25 25

-5 30 -25

1 -6 5 0


Our second degree polynomial is of the sort. x^2-6x+5=0

This of couse can easily be factored! into (x-1)(x-5)

Therefore our zeros of the polynomial function are -1,-5, 1 and 5.

So our linear factors are (x-1)(x+1)(x-5)(x+5)= x^4-26x^2+25.

I know what some of you are thinking.....So what does this have to do with square root of f(x). It has a lot to do with it! First we need to factor the polynomial and since that all square roots are postive we need to set the linear factors greater than or equal to zero.

So (x-1)(x+1)(x-5)(x+5)=0

Set it equal to zero, we find the critical points 1,-1,5,-5

therefore we have the intervals of:

(-inf, -5) U (-5,-1) U (-1,1) U (1,5) U (5,inf).

+ - + - +


We find where the graph is greater or equal to zero. It is with 3 intervals!

So our answer to finding the square root of f(x) is the intervals:

(-inf, -5) U (-1,1) U (5,inf)

Punch this function into a calculator---- very interesting piecewise function.

How about a function like this:

x^4+6x^3+11x^2+6x+1. The square root function is not much different than the original and it is hard to find the zeros to this function. Do you know why? By Desecrates rule of signs there are no sign variations therefore there are zero positive rational zeros. To find the negative rational zeros you try f(-x) and f(-x)= (-x)^4+6(-x)^3+11(-x)^2+6(-x)+1. Hence f(-x) is x^4-6x^3+11x^2-6x+1. There are 4 variations of this. So there are either 4 or 4-2=1 or 4-4= 0 variations. I plugged this equation into my voyage 200 calculator and the zeros are irrational zeros. So if anyone knows how to find the zeros to this function let me know. There are only 2 and they are both irrational.

My Math Story

Perhaps some of you guys are wondering how I came to be so "in love" with math? Well, it happens to be a relatively long story. I cannot say I have all the answers-- answers to questions that are known to human-kind about mathematics or even to a very advanced level, but I can say that I progress toward the goal of knowing and doing math and to the best of my ability. I have taken tests in the past and my math skill has always been "above average." This doesn't mean that there are some that can do math and some that cannot. Some may have the tendency to have trouble, I think a lot of it has to do with teachers and how good they are to be honest. In eighth grade math was one of my worst subjects, and in high school it turned into one of my favorites.

When I say teachers have an influence I mean for better or for worse. The teachers that have influenced me for the good are professors at Lansing Community College. The first teacher that rendered good influence was Nan Jackson. I had her for College Algebra I. She always wanted us to think out the problem and draw connections in our minds. The second teacher that was of good influence was Rob Stiegemeyer. He would sit upstairs at the LCC LSARC tutoring desk and help many-- anyone who needed help and was a very good teacher. To be honest, I think sometimes we take far too much advantage of such tutors, having them work out the whole problem instead of bringing our ideas to them also. The third and final teacher that was of good influence to me was Jing Ling Wang. After learning fractals in Nan Jackson's class and having her as a guest speaker in our Geometry class, I wanted to take her Calculus class. To these teachers I dedicate this blog.

One of my pursuits or goals in mind is to have a book written. I am thinking perhaps when I accumulate enough posts on this blog, I will make it into a book and publish it. I use the mark-up language LaTeX to write my book, so I better start using it instead of using ^ etc! That will be another adventure in itself!

Currently I am trying to get students to tutor in my own home. I CAN use what I've learned even though I haven't got a Bacherlor's degree! Use what you have learned thus far, and you can go further!

Calculating Square Roots by Hand Part 2

Here is a challenge for you fellow math fans out there... I am a avid math book collector. I have math books dating from the 1860's to the present day! Here is a excerpt from a math book dated 1906 on how to compute square roots by hand. Before this explanation, there is a very large section on computing square roots on polynomials. Not easy ones like (x+y)^2 but for example here is one:

Find the square root of:
20 x^2 - 22x^3 + 1 +28 x^4 +9x^6 - 8x - 12x^5

Isn't that nuts? I've never seen a problem like this in a text book that I've used. But anyways here is an explanation how to compute square roots by hand:

We then have the following rule for finding the square root of an integral perfect square.

Separate the number into periods by pointing every second digit beginning with the units' place.
Find the greatest square in the left-hand period, and write its square root as the first digit of the root; subtract the square of the first root digit from the left hand period, and to the result annex the next period.
Divide this remainder, omitting the last digit, by twice the part of the root already found, and annex the quotient to the root, and also to the trial-divisor.
Multiply the complete divisor by the root-digit last obtained and subtract the product from the remainder.
If other periods remain. proceed as before. doubling the part of the rot already found for the next trail divisor.

Could anyone explain why this procedure works? Nan, Rob, or Jing, or anyone? I am planning on studying on calculating square roots on large polynomials and be able to teach and explain. There is also a very large section in the book on cube roots of polynomials----- fun fun fun!!!!! =)

Why is a Number divided by 0 undefined?

There are two answers that I am aware of to this question. (If anyone else knows any other answer to this question, please let me know by posting a comment).

The first is a pretty easy explanation. First of all we want to distinguish between indeterminate and the undefined. Indeterminate means that we cannot determine the answer because there are too many answers to the question. (an example would be 0/0.) Because we know that division is the opposite of multiplication, we can use multiplication to get our answer. 0/0 means:

___ x 0 = 0 The answer to this question is ANYTHING. Any number times 0 is 0. So that is why it is indeterminate.

On the other hand, there is also the undefined. We can apply the same rule-- multiplication being the inverse of division to find our answer. 0 x ____ = a

"a" stands for a constant-- any constant, may it be 6, or 7, or 1, or 2..... but not 0. You cannot times 0 by any number and get a constant a. This is the simple reason why we cannot divide by zero.

Now for the more complex reasoning..... 1/0 could be infinity or negative infinity. If we divide 1 by a tiny positive number, the answer will be a big positive number. If we divide 1 by a tiny negative number, the answer will be a big negative number. In other words, as our denominator gets closer to 0 from the right we're going to infinity, as it gets closer to 0 from the left, we're going to negative infinity. That's why we let 1/0 be undefined. This concept is explored in Algebra with asymptotes, and in Calculus with the idea of Limits in more detail.

Julie Code

Calculating Square Roots by Hand

Here is an interesting thing to consider.... The square root of 100 is 10; of 10000 is 100; etc. Hence the square root of a number between 1 and 100 is between 1 and 10; the square root of a number between 100 and 10000 is between 10 and 100; etc. A round square number is a number that ends in 0 and is square. We can use these types of numbers to calculate other square roots of numbers.

Here is a pattern to use,

Number ends in 0 then its square root ends in 0 Iff it is a square number
Number ends in 1 then its square root ends in 1 and 9
Number ends in 2 then there is no square root--- the number is irrational
Number ends in 3 then there is no square root--- the number is irrational
Number ends in 4 then its square root ends in 2,8 Iff it is a square number
Number ends in 5 then its square root ends in 5 iff it is a square number
Number ends in 6 then its square root ends in 4 or 6 iff it is a square number
Number ends in 7 then there is no square root-- the number is irrational
Number ends in 8 then there is no square root--- the number is irrational
Number ends in 9 then its square root ends in 3, or 7 iff it is a square number

Here are the round square numbers......

10 x 10 = 100 therefore its square root is 10
20 x 20 = 400 therefore its square root is 20
30 x 30 = 900 therefore its square root is 30
40 x 40 = 1600 therefore its square root is 40
50 x 50 = 2500 therefore its square root is 50
60 x 60 = 3600 therefore its square root is 60
70 x 70 = 4900 therefore its square root is 70
80 x 80 = 6400 therefore its square root is 80
90 x 90 = 8100 therefore its square root is 90
100 x 100 = 10000 therefore its square root is 100

Therefore if the number is between 81 and 10,000 then the square root will be only a two digit answer. Let us do some problems with this.

Lets do the square root of 2025. It has to be between the square root of 1600 and 2500 because the square root of 1600 is 40 and the square root of 2500 is 50 and the number 2025 is between the two round square numbers of 1600 and 2500. It has to start with a 4 logically and cannot start with a 5 because it cannot be above 50. Since it ends in a 5 the square root has to end in a 5. So the answer is 45.

Lets do the square root of 2187. STOP right there! WE see that it ends in a 7 therefore we cannot write it as ratio of two integers..... as whole numbers are rational.

Lets try to do the square root of 7529. It is between the round square numbers of 6,400 and 8,100 so the square root would have to be between 80 and 90. Since it ends in a 9 the answer would have to be a two digit answer starting with an 8 and ending in either a 3 or 7. If you do the math 83 x 83 or 87 x 87 you will NOT get 7529. We can therefore conclude that the answer is irrational.

This is how we calculate square roots in our heads or on paper. We will next try to take square roots of polynomials in our next lesson--- I found this in a book dated 1906-- very interesting.... until then ta ta!

Julie Code