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Archive for the ‘Algebra’ Category

Percents Defying Intuition


Question related to percents of increase and decrease … we know what happens if you increase an original quantity by adding x% of it, then subtracting that same x%: you’ll wind up with the quantity you started with. That is to say, the value of the original quantity will stay the same.

Now let’s pose the same question with regard to multiplication and division. Suppose you take an original quantity and increase it by x%. Then you turn around and decrease the new quantity by x%?

Will the new amount be: a) the same? b) greater? c) less? or d) might it depend on the value of x? What if you reverse the order … you first decrease the quantity by x%, and then you turn around and increase that new quantity by x%. Will the amount you wind up with be the same or different than what you got the other way around?

Many students intuitively think the final result will be the same as the original quantity. That makes this concept fun to teach; math lessons always are more fun when the results go against intuition.

We’ve already said that the amount of the increase and decrease is x%.

Now to keep the math as simple as possible, we’ll just use 1.00 as the original quantity. That way all values that we wind up with will automatically be in percent form.

When we increase our original quantity, 1.00, by x percent, we multiply it by 1.00 + x% = 1.00 + x/100 = 100/100 + x/100 = (100 + x)/100.

Similarly, when we decrease the original quantity, 1.00, by x percent, we multiply it by 1.00 – x% = 1.00 – x/100 = 100/100 – x/100 =(100 – x)/100.

So, then what are the steps performed in the problem? In the first way described, we 1st) increase the original quantity by x%, then 2nd) starting with the increased value, decrease it by x%. Let’s do that now.

1st) 1.00 [times] (100 + x)/100 = (100 + x)/100

2nd) (100 + x)/100 [times] (100 – x)/100 = (100 + x)(100 – x)/100^2 = (100^2 – x^2) / 10,000 = (10,000 – x^2) / 10,000.

We can draw a number of conclusions from this final algebraic expression, boldfaced.

1) It represents the quantity we are left with after increasing, then decreasing a quantity of 1.00, by x%.

2) Since this was done in a general way, this expression serves as a formula to predict the new value for any situation where an original quantity gets increased or decreased by the same percent.

Example: if $2,000 gets increased by 3%, then decreased by 3%, the amount left would be given by this:
2000 [times] (10,000 – 3^2) / 10,000 = 2000(10,000 – 9)/10,000 =
2000(9991/10,000) = 1998.2

3) Since this expression by which the original value gets multiplied does not changed, no matter the original value, no matter the percent, it behooves us to deeply understand this expression:
(10,000 – x^2) / 10,000

4) First, let’s notice that this expression does not change whether we increase the original value first, then decrease — or if we switch the order by decreasing the original value first, then increasing the result. We can see this is true because the numerator of the expression: (10,000 – x^2) arises as the product of (100 + x) and (100 – x). The commutative property guarantees that the order of the factors does not change their product.

So the point here is that it doesn’t matter whether we first increase the original quantity, then decrease it, or if we first decrease the original quantity, then increase it; we’ll get the same result each time.

5) Now let’s look at the relationship between the value of the percent, x, and the outcome, the change in the original quantity.

From the expression we see that as x^2 gets larger, the numerator,
10,000 – x^2, will get smaller. And of course, the value of x^2 increases as the value of x increases.

So in a transitive way of thinking, as the percent, x, that we increase and decrease by gets larger, the ultimate decrease of the original value gets larger. Or, slightly differently, as the percent of change we put the value through increases, the original quantity will end up decreasing more at the end.

Here’s a table that shows the relationship between the x-value, the percent of increase and decrease, and the percent of the original value you’re left with after the increase and decrease are carried out.

This is the first time I’ve explored this topic. So feel free to share your thoughts, insights on it.



Factoring Trick: How to Flawlessly Factor any “Difference of Two Squares” Binomial


If you’re staring at two terms you need to factor, but feel like a deer looking at the headlights of an oncoming semi, here’s a way to leap to safety!

It’s called the “Difference of Two Squares” trick.

High-Octane Boost for Math

It requires four simple steps.

  1. Figure out if each of the terms is a “perfect square.”
  2. If so, take the square root of each term.
  3. Put each square root in its proper place inside two (    ).
  4. Put a + sign inside the first (   ), and put a – sign inside the second (   ).

Let’s do an easy example. Suppose the terms you’re looking at are these:
x^2  – 9

Let’s go through the 4 steps together.

  1. Figure out if each term is a “perfect square.”

    So, what does it mean for a number or term to be a “perfect square”?  It means that you get the number or term by multiplying a number or term by itself. For example, 16 is a perfect square because you can get 16 by multiplying 4 by itself:  4 x 4 = 16.

    So when we look at our two terms, x^2 and 9, we notice that both
    are perfect squares.
    9 is just 3 times 3.
    And in the same way, x^2 is just x times x.

  2.  Take the square root of each term.
    The square root of x^2 is just x.
    And the square root of 9 is just 3.

  3. Put each square root in the proper place inside two sets of (    ).
    We put the square root of the term that was positive first, and the square root of the term that was negative second.Since the x^2 was the positive term, we put its square root, x, first inside each
    (   ).  So far, that gives us:  (x    ) (x     )

    Since the 9 was the negative term because it had the negative sign in front of it: – 9, we put its square root, 3, second inside each (   ). So our (   )s now look like this:  (x   3) (x   3)

  4. Finally, we just need to put in signs that connect the terms inside
    the (    )s.

    That’s easy. We put a + sign inside one (    ), and we put a – sign
    inside the other (    ).
    I prefer to put the + inside the first (   ), but it really doesn’t matter.The final factored form, then, looks like this:  (x + 3) (x – 3)
    That’s all there is to it.

Now try these problems for practice.

           a)  x^2 – 16
           b)  x^2 – 100
           c)   x^2 – 121
           d)   x^4 –  16x^2
           e)   49x^8 – 144y^12

Answers:

           a)   x^2 – 16   =  (x + 4) (x – 4)
           b)  x^2 – 100  = (x + 10) (x – 10)
           c)   x^2 – 121  = (x + 11) ( x – 11)
           d)   x^4 –  16x^2  = (x^2 + 4x) (x^2 – 4x)
           e)   49x^8 – 144y^12  = (7x^4 + 12y^6)(7x^4 – 12y^6)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

How to Divide ANY Number by a Radical — Fast!!! (Math “hack” w/ full explanation)


 

Here’s a super-quick shortcut for  DIVIDING ANY NUMBER by a RADICAL. 

Note: I’m using this symbol () to mean square root.
So √5 means the square root of 5;  √b means the square root of b, etc.
 And … if you want to learn why this “hack” works, see my explanation at the end of the blog.

This “hack” lets you mentally do problems like the following three. That means you can do these problems in your head rather than on paper.

     a)  12 / √3 

     b)  10 / √2

     c)  22 / √5

Here are three terms I’ll use in explaining this “hack.”

In a problem like 12 divided by √3, which I write as:  12 / √3,

     12  is  the dividend,

     3  is  the number under the radical,

     √3  is  the radical.

The “Hack,” Used for  12 / √3:

  1.  Divide the dividend by the number under the radical.
    In this case, 12 / 3  =  4.
  2. Take the answer, 4, and multiply it by the radical.
    4 x √3  =  4√3

  3. Shake your head in amazement because that, right there, is the ANSWER!

Another Example:  10 / √2

  1.  Divide the dividend by the number under the radical.
    In this case:   10 / 2  =  5
  2. Take the answer you get, 5, and multiply it by the radical.
    5 x √2  =  5√2.  (Don’t forget to shake head in amazement!)

Third Example:  22 / √5

  1.  Divide dividend by number under the radical.
    In this case,  22 divided by 5 = 22/5  (Yep, sometimes you wind up with a fraction or a decimal; that’s why I’m giving an example like this.)
  2. Take the answer you get, 22/5, and multiply it by the radical.
    22/5 x √5 =  22/5 √5.  [Note: the √5 is in the numerator, not
    in the denominator. To make the location of this √5 clear, it’s best
    to write the answer:  2√5 / 5].


NOW TRY YOUR HAND by doing
these PRACTICE PROBLEMS:

a)   18 / √3  

b)   16 / √2  

c)   30 / √5  

d)   10 / √3  

e)   12 / √5

– – – – – – – – – – – – – – – – – –

ANSWERS:

a)   18 / √3  = 6√3

b)   16 / √2  = 8√2

c)   30 / √5  = 6√5

d)   10 / √3  = 10√3/3

e)   12 / √5  = 12√5/5

– – – – – – – – – – – – – – – – – –

WHY THE “HACK” WORKS:

It works because we rationalize the denominator of a fraction whenever the denominator contains a radical. Here’s the “hack” in general terms, with:

     a  =  the dividend,

     b  =  the number under the radical,

     √b  =  the radical.

a / √b

=   a
    √b

=   a     √b    =   a √b
    √b   √b            b

Notice: we started with:  a / √b.

And keeping things equal, we ended up with  a √b / b.

This shows that the “hack” works in general. So it works in all specific cases as well!

– – – – – – – – – – – – – – – – – –

Final note: the number under the radical is called the radicand. But that term is so close to the term radical that I thought it would be less confusing if I just called this the number under the radical. I hope you are not offended.

 

 

 

 

 

 

Algebra Mistake #5: How to Combine a Positive and a Negative Number without Confusion


So, you’d think that combining a positive number and a negative number would be a fairly straightforward thing, huh?

Well, unfortunately, a lot of students think it’s easy. They think it’s too easy. They think there’s one simple rule that guides them to the very same kind of answer every time. And that’s exactly where they get into trouble.

The truth is that combining a positive and a negative number is a fairly complicated operation, and the sign of the answer is dependent on a nmber of factors.

This video reveals a common mistake students make when tackling these problems. it also shows the correct way to approach these problems, using the analogy of having money and owing money to make everything make sense.

So take a look and see if this explanation doesn’t end the confusion once and for all.

And don’t forget: there are practice problems at the end of the video. Do those to make sure you’ve grasped the concept.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Algebra Mistake #4: How to Combine Negative Numbers without Confusion


Here’s a common mistake, and a very understandable one, too. Students need to combine two negative numbers, and they, of course, wind up with an answer that’s positive. Why? Because, they’ll say — pointing out that you yourself have told them this —  “Two negatives make a positive!”

This video gets to the root of this common misunderstanding by helping students understand exactly when two negatives make a positive, and when they don’t.

 

Make sure you watch the whole video, as there are practice problems at the end, along with their answers.

 

 

 

 

 

 

 

 

 

“Algebra Survival” Program, v. 2.0, has just arrived!


The Second Edition of both the Algebra Survival Guide and its companion Workbook are officially here!

Check out this video for a full run-down on the new books, and see how — for a limited time — you can get them for a great discount at the Singing Turtle website.

 

Here’s the PDF with sample pages from the books: SAMPLER ASG2, ASW2.

And here’s the website where you can check out the books more fully and purchase the books.

 

 

 

 

 

 

 

How to Factor Trinomials with Understanding!


This video shows the fastest and easiest way I know of for factoring quadratic trinomials. Give it a watch and see if you agree.