This post is, once again, a summary of a longer post authored by me on my own blog. My blog covers a lot of areas, including Vedic Mathematics. If you are interested in reading my thoughts on other topics, please feel free to visit my blog and post comments on the other articles you find there also! Thank you!
In the previous three lessons, we saw how it is possible to solve certain quadratic equations without using the quadratic formula. In fact, these are equations which we don't even have to expand out to the standard form for applying the quadratic formula. In this lesson, we will deal with a few more types of quadratic equations. Once again, these qu! adratic equations are not expressed in standard form, so these solution techniques save us the trouble of both converting the equations to standard form, and also the trouble of then applying the quadratic formula to solve them.
The first type of non-standard quadratic equation was already dealt with in this earlier lesson in the context of the Sunyam Samyasamuccaye sutra. These are equations in which the sum of the terms in the numerator is the same as the sum of the terms in the denominator, and the difference between numerator and denominator of each term is the same on both sides of the equal to sign. An example of such an equation is:
(2x + 1)/(3x + 2) = (5x + 2)/(4x + 1)
You will immediately notice that the sum of the denominators of both terms on either side of the equal to sign i! s 7x + 3. Similarly, the sum of the numerators of both terms o! n either side of the equal to sign is also 7x + 3. The Sunyam Samyasamuccaye sutra says that in that case, this sum can be set to zero for one solution of the equation. Thus, 7x + 3 = 0, giving us x = -7/3 as one solution. We also see that the difference between the numerator and denominator on either side of the equal to sign is x + 1. Application of the sutra then tells us that this difference can be set to zero for the other solution to the equation. Thus, x + 1 = 0, giving us x = -1 as the other solution.
As we saw in the earlier lesson, this is very useful in solving equations that are expressed as below:
(ax + b)^2 = (cx + d)^2
where a is not equal to c.
We can then rewrite the equation mentally, as below:
(ax + b)/(cx + d) = (cx + d)/(ax + b)
We know that in this form, the sunyam samyasamuccaye sutra is applicable in b! oth its meanings, giving us two solutions of the quadratic equation immediately. One meaning tells us that (a + c)x + b + d = 0, which gives us x = -(b + d)/(a + c). The other meaning tells us that (a - c)x + b - d = 0, giving us x = (d - b)/(a - c).
For instance, consider the equation (3x + 5)^2 = (2x + 1)^2. We can spend a lot of time expanding out the terms, collecting them on one side and deriving an equation that is expressed in standard form, and then apply the quadratic formula to it. Or, we could recognize that this is a case where the sunyam samyasamuccaye sutra is applicable, and derive the two roots mentally as x = -6/5 and x = -4!
Now consider another equation of the following type:
a/(x + a) + b/(x + b) = c/(x + c) + d/(x + d)
Also assume that a + b = c + d. In that case, the sunyam samyasamuccaye sutra tells us that one of the roots is x = 0, and the other ro! ot is derived by setting (x + a) + (x + b) = 0. Why? Consider ! the foll owing manipulation of the first term, a/(x + a).
a/(x + a) = 1 - x/(x + a)
By similar manipulation of all the terms, we can rewrite the equation as:
1 - (x)/(x + a) + 1 - (x)/(x + b) = 1 - (x)/(x + c) + 1 - (x)/(x + d)
Cancelling out the 1's from both sides of the equation, we get:
x/(x + a) + x/(x + b) = x/(x + c) + x/(x + d)
This then tells us that x = 0 because it is a common factor in all terms of the equation (from this previous lesson). We can then remove the x from the numerator of the equation leaving us with:
1/(x + a) + 1/(x + b) = 1/(x + c) + 1/(x + d)
Since a + b = c + d, the sum of the denominators on the left-hand side of the equation is the same as the sum of the denominators on the right hand side of the equation. Equating this sum to zero gives us the other root (this result is ! from this previous lesson). Thus, setting 2x + a + b = 0 gives us x = -(a + b)/2 as the other root of the equation.
Consider the equation below:
1/(x + 1) + 4/(x + 4) = 2/(x + 2) + 3/(x + 3)
We find that this follows the pattern we have dealt with just now, and also 1 + 4 = 2 + 3. We can then say, on sight, that the solutions to this equation are x = 0, and x = -2.5.
The next special type of equation we will consider is very similar to the merger type of equation we dealt with in this previous lesson. However, the sum of the numerators on the left hand side of the equation is not equal to the numerator on the right hand side, as required for the merger operation. Instead, consider the equation below:
a/(x + b) + c/(x + d) = e/(x + f)
where a/b + c/d = e/f.
When the above co! ndition is satisfied, then one of the roots is zero, and the other root can be derived by merger. Let us see how.
a/(x + b) + c/(x + d) = e/(x + f) can be manipulated to
a/b - (ax/b)/(x + b) + c/d - (cx/d)/(x + d) = e/c - (ex/f)/(x + f)
Since a/b + c/d = e/f, we can cancel those terms out of the equation above, giving us:
(ax/b)(x + b) + (cx/d)/(x + d) = (ex/f)/(x + f)
Now, since x is a common term in all three numerators, one of the roots is x = 0. Removing the x from the numerator of all three terms gives us:
(a/b)/(x + b) + (c/d)/(x + d) = (e/f)/(x + f)
Once again, since a/b + c/d = e/f, the above equation can be solved by merger operation easily. Consider the example equation below:
6/(x + 3) + 4/(x + 2) = 20/(x + 5)
We apply the test above, and find that 6/3 + 4/2 = 20/5. Therefore, we can set one root to x = 0, and then simplify the equation as below:
2/(x + 3) + 2/(x + 2) =! 4/(x + 5)
The new numerators are a/b, c/d and e/f as worked out in the algebraic equation. We can then solve this by merger by setting p = 2, q = 2, a = 3, b = 2 and c = 5. The solution is x = -13/5. Thus the two roots of the given equation are x = 0 and x = -13/5.
Before I leave, I want to touch upon a set of equations that may be presented in standard form, but which may be easier to solve by converting to a special type. These are equations of the type ax^2 + bx + c = 0 where a and c are both 1. Thus the equations have the form x^2 + bx + 1 = 0. It is easy to see that these equations can easily be converted to the following form:
x + 1/x = -b
As we saw in the previous lesson, this kind of equation is easy to derive approximate roots for mentally. For instance, if you are faced with the equation x^2 -10x + 1 = 0, you can instantly es! timate the roots to be x = 9.90 and x = 0.10. The exact roots ! turn out to be 0.1010 and 9.8990!
Similarly, if the equation in standard form is presented as x^2 + bx - 1 = 0, then it is easy to see that this can be converted to x - 1/x = -b. This, again, can be solved mentally as we saw in the previous lesson to derive approximate roots. Thus, if asked to solve x^2 + 10x - 1 = 0, we can quickly say that the roots are approximately 0.10 and -10.10. The exact roots turn out to be 0.0990 and -10.099!
Hope this lesson (combined with the previous lessons on quadratic equations) has provided you with a little more insight into how certain quadratic equations can be solved without going through the trouble of expressing them in standard format and applying the quadratic formula. We have also dealt with a couple of speci! al cases of equations that are expressed in standard form being more amenable to mental solution than application of the quadratic formula. Being able to recognize the applicability of these techniques is very important to realize the time and labor savings associated with these techniques. Please be sure to practice these techniques so that their application comes as naturally as the application of the quadratic formula! Good luck, and happy computing!!
quadratic equation
Tuesday, August 31, 2010
Microsoft Access Tips & Tricks: Probability Density (pdf)
Microsoft Access (or any other relational database product that supports SQL) can be used to produce a table of distinct values in a given column along with the probability density of each of these values using SQL.
If you are interested, you can find my earlier lessons on finding the median, the mode, the geometric and harmonic means, ranking every row in a query, selecting random rows out of a table, calculating running sums and averages, calculating running differences, and creating histograms out of given data.
Technically speaking, since all data in a database consists of discrete values, the measures we are calculating in this post should correctly be referred to as Probability Masses. A Probability Mass Function (pmf) is a functi! on that gives the probability that a discrete random variable ! is exact ly equal to some value. If you actually consider the data in your database to be samples from a truly continuous distribution and want a better measure of the probability of small subranges of it, you may be interested in creating a histogram out of the data for further analysis.
In this post we will assume that we have a table myTable that contains a field myField with several different values. The probability mass of any given value in myField is the number of times that given value occurs in myField divided by the total number of rows in myTable. The sum of all probability masses will always equal one.
This is a pretty basic definition, and as such, it is a good place to begin. Later we will see how to deal with fields that have NULL values that one may want to exclude from the calculations and/or output.
Given the definiti! on above, it is easy to see that the SQL below will produce a table of probability masses:
select myField, count(*)/(select count(*) from myTable) from myTableThe sub-query in the above SQL provides us the denominator of the probability mass by providing us the count of the total number of rows in myTable. The count(*) in the outer query counts the number of occurrences of a given value of myField in the column. The group by clause produces one row of output for each distinct value of myField, and the order by clause arranges these rows of output in ascending order of value of myField.
group by myField order by myField
The order by clause can be modified to arrange the rows in the output by any other measure, such as the probability mass of each value of myField also.
What happens when there are NULL values in myField? You will see that your output will consist of one row for the NULL values along with a probability mass for it (it wi! ll be the top row if your order by clause is the same as the o! ne used in the SQL provided here). If you do not want that row in your output, you have to explicitly exclude that row from the output in your SQL as below:
select myField, count(*)/(select count(*) from myTable) from myTableThe query above uses a where clause to filter out NULL values in myField. Now, in the output, you will notice that there is no row with a NULL value of myField. However, the probability masses of the values that are included in the output may not add up to one. The fraction that is missing (if any) is the fraction of NULL values that we have excluded from the output.
where myField is not NULL
group by myField order by myField
In fact, one can modify the where clause of the above SQL to exclude not just NULL's, but any other values or ranges we want from the output (or, we can explicitly list the values or ranges we want to include in the output too). Moreover, the where clause can be based on any columns in the table, no! t just myField. However, since the sub-query still includes these values in counting up the rows in the table, our denominator never changes, thus resulting in the probability masses not adding up to one.
To get the probability masses to add up to one, we need to adjust the denominator of our calculations to only include rows that are included in the outer query. This may be valid in some contexts (for instance, when we eliminate NULL values in myField), but may not be valid in some other contexts. So, one has to use caution and statistical judgment in determining whether the probability masses that are being calculated should actually add up to one or not.
To adjust the denominator so that it also includes only those rows that are considered by our outer query, there are two options. The first and easiest option is to include the outer query's where clause verbatim in the subquery. Thus our SQL could look something like this:
select myF! ield, count(*)/(select count(*) from myTable where C) from myT! ableThe condition in the outer query is denoted here by C, and it is used again in the subquery so that the number of rows in the outer query (before the grouping occurs) is the same as the number of rows in the subquery. This will make the probability masses add up to one.
where C
group by myField order by myField
If you are only eliminating NULL values from myField in the calculations, there is a much more elegant way of removing the rows where myField is NULL from the count in the subquery:
select myField, count(*)/(select count(myField) from myTable) from myTableThe subquery in this case takes advantage of the fact that count() with the name of a field provided only produces the count of rows with non-NULL values in that field. Thus the use of count(myField) automatically eliminates the rows in which myField is NULL, which is the same restriction as that imposed b! y the where clause of the outer query!
where myField is not NULL
group by myField order by myField
Hope this post has been helpful in solving any problems you might have had with calculating probability masses out of your data in Access. If you have any problems or concerns with the queries in this lesson, please feel free to let me know by posting a comment. If you have other questions on Access that you would like me to address in future lessons, please feel free to let me know through your comments too. Good luck!
calculating probability
PreAlgebra Quiz1 Linear
Hope this helps. See you tomorrow!
Feel free to watch as many times as you like.
Online pre algebra equations
What is Multiplication?
Can be a tricky question, this.
Primary schoolchildren are taught that it is repeated addition, which makes a lot of sense.
4 x 4 =
4 + 4 + 4 + 4 =
16
Four lots of four/groups of/sets of.
But multiplication is also a scaling quantity. It ratchets things up massively quickly. If you type 2x2 into your calculator, then keep multiplying every answer by 2, the calculator will quickly run out of digits. To compare this with addition misses the point that multiplication is the centre of all geometric operations.
It does through the squaring and cubing and so on effect. If you take four centimetres, and for each of these four, you add another four, you get 4 x 4. It is known as squaring because the Greeks used to see it geometrically as the way to find the area of a square. For each centimetre across, there are four up (and vice versa).
x2 is a powerful, recurring idea, which has ! its role in pretty much the entire universe. (E = mc2)
It's also worth noting the effect of multiplication by a fractional quantity.
4 x 1/4 will obviously increase the fraction, but it will have a decreasing effect on the whole number. Unlike multiplication by two whole numbers, where the answer is greater than either, in this operation, the answer will always be smaller than one of the inputted values.
adding and subtracting polynomials calculator
Primary schoolchildren are taught that it is repeated addition, which makes a lot of sense.
4 x 4 =
4 + 4 + 4 + 4 =
16
Four lots of four/groups of/sets of.
But multiplication is also a scaling quantity. It ratchets things up massively quickly. If you type 2x2 into your calculator, then keep multiplying every answer by 2, the calculator will quickly run out of digits. To compare this with addition misses the point that multiplication is the centre of all geometric operations.
It does through the squaring and cubing and so on effect. If you take four centimetres, and for each of these four, you add another four, you get 4 x 4. It is known as squaring because the Greeks used to see it geometrically as the way to find the area of a square. For each centimetre across, there are four up (and vice versa).
x2 is a powerful, recurring idea, which has ! its role in pretty much the entire universe. (E = mc2)
It's also worth noting the effect of multiplication by a fractional quantity.
4 x 1/4 will obviously increase the fraction, but it will have a decreasing effect on the whole number. Unlike multiplication by two whole numbers, where the answer is greater than either, in this operation, the answer will always be smaller than one of the inputted values.
adding and subtracting polynomials calculator
Chapter 5 Review Solutions
Here are the solutions for the Chapter 5 Review:
Calculators are allowed tomorrow. Please remember to bring your review sheet, as that acts as your admission ticket to take the test. You will also need a pencil. I will provide you with any scratch paper you might need. Good luck!
algebra 2 answers
Calculators are allowed tomorrow. Please remember to bring your review sheet, as that acts as your admission ticket to take the test. You will also need a pencil. I will provide you with any scratch paper you might need. Good luck!
algebra 2 answers
Pythagorean auctions and other Pythagorean excitement for middle school teachers at Bard College next week!
As I said, why should students have all the fun?
Next Monday, I will be leading a session on Pythagorean auctions at the New York City Math Teachers Circle summer workshop for middle school mathteachers at Bard College.
The inspiration for running math auctions in my session came from an idea I first read about in the book Mathematical Circles (Russian Experience) by Dmitri Fomin, Se! rgey Genkin, and Ilia Itenberg. Further encouragement came from a very enthusiastic presentation by Anna Burago at MSRI's Great Circles 2009 conference. Anna has many years of experience with math circles, both as a student in her native St. Petersburg and now as a lead teacher for the Northwest Academy of Sciences math circle. Here is her description:
Mathematical Auction is an exciting team mathematical contest. It combines mathematics with the elements of gambling and psychological thriller. The game starts from a period of problem solving which is followed by a round of trading. The teams bid, scheme and strategize for the rights to ! present the solutions to the problems. Mathematical Auction is! one of the favorite competitions in the Math Circles that I run.As an economist, of course, I was intrigued by the idea of a mathematical auction, so I had to try the idea out in our math circle. Since Anna said that the auctions are especially popular with middle schoolers, we ran some math auctions at our Albany Area Math Circle middle school math circle meetings last spring--and they were indeed very well received by the students. I hope the teachers at next week's workshop will also enjoy using math auctions to explore some interesting Pythagorean mathematics.
Here are the rules for a Math Auction:
1) The auctioneer hands the teams a set of problems--about five or six problems is considered ideal. But the problems are a special kind--Russian math circle leaders call them "research problems" because they allow possibilities for partial solutions or intermediate answers which can gradually lead to a fin! al result.
2) Teams are given a certain amount of time to work on the problems.
3) When that time is up, the auction begins. Each team starts with a certain amount of fictional currency at the outset of the auction. In honor of Pythagoras, we'll use drachmas as our fictional unit of currency in next week's workshop, so each team will have 1,000 drachmas to use in their bidding.
4) The teams then bid for the right to present a solution to each problem. The team that submits the highest bid in the initial round of the auction gets the right to present their solution. After they present their solution, the bidding reopens so that other teams who believe they can improve on that solution may do so. The team with the strongest solution to the problem wins the value of the problem. Bonus drachmas may be awarded for cool discoveries made alone the way.
The focus of t! he summer teacher workshop is the Pythagorean theorem, so all ! the prob lems we will be auctioning off next Monday afternoon will use that theorem in a variety of fascinating ways--many of the auction problems will connect to the presentations that will be given by other speakers later in the week.
Here is the list of workshop presentations scheduled at the Bard math teacher summer workshop next week:
Auctions of All Things Pythagorean: Spirals, Trees, Triples, Twins, Quads, Networks, and Outcastes, Mary O'Keeffe (Albany Area Math Circle)
Showing of The Theorem of Pythagoras (movie created by Tom Apostol's Project Mathematics! at Caltech with awesome animation from MacArthur "genius" prize winner Jim Blinn )
Primitive Pythagorean Triples Sheila Krilov (Hunter College High School teacher and MATHCOUNTS coach)
Heronian Triangles David Hankin (former chair of the AIME Com! mittee and veteran teacher/math department chair Hunter College High School)
Showing of The Proof (movie about Andrew Wiles proof of Fermat's Last Theorem)
Pythagoras in Spherical and Hyperbolic Geometry, Jim & Maria Belk (Bard College faculty)
Eight Different Proofs of the Pythagorean Theorem Gary Rubinstein (Stuyvesant High School faculty)
Fermat n=4 and Some Interesting Open Problems in Number Theory Lauren Rose (Bard College faculty)
Primitive Pythagorean Triples via Unique Factorization John Cullinan (Bard College faculty)
Almost Pythagorean Triples, and Almost Isosceles Pythagorean Triples: Connecting to Pell's Equation, Japheth Wood (Bard College faculty)
All of this mathematical excitement (as well as music and film festivals) will take place on Bard's campus next week. The area near Bard, overlooking the Hudson River, is a spectacularly beautiful setting at this time of year (word has it that Chelsea Clinton will be getting married nearby shortly after the math circle teacher workshop ends!)
There may be still room for a few more math teachers to attend the workshop--and it is even possible to arrange for continuing education credit. Please pass the word along to any middle school math teachers you know who would enjoy this experience!
the area of a circle
Solving Quadratic Equations
Here are a few sites that may help you when solving quadratic equations:
Algebra quadratic equations
- by factoring
- by finding square roots
- by completing the square
- by Quadratic Formula
- by graphing
- extra examples
Algebra quadratic equations
Subscribe to:
Posts (Atom)