Kiss those Math Headaches GOODBYE!

Archive for the ‘Practice Problems’ Category

Factor Quadratic Trinomials, Part 2


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

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Algebra Mistake #2: Does a x a = 2 x a?


Now that you’ve gotten a taste for the benefits of analyzing algebraic mistakes, it’s time to explore a second common mistake. This one is so common that nearly every student commits it at least once on the road to algebra success.

As you watch the video, notice how by thinking hard about two expressions, we can think this mistake through to its very root, thus discovering the core difference between two similar-looking algebraic expressions.

And along the road, we’ll learn a general strategy for decoding the meaning of algebraic expressions. What I like about this strategy is that you can use it to understand the meaning of pretty much any algebraic expression, and you’ll see that it’s not a hard thing to do. In fact, it just involves using numbers in a nifty way.

Best of all, students usually find this approach interesting, convincing and even a bit fun. So here goes, Common Algebra Mistake #2 …

 

How to Remove (“Unpack”) Algebraic Terms from Parentheses


As you’re probably aware, I’m a big believer in using stories to bring math to life. Especially when you’re teaching tricky concepts, using a story can be the “magic switch” that flicks on the light of understanding. Armed with story-based understanding, students can recall how to perform difficult math processes. And since people naturally like stories and tend to recall them, skills based on story-based understanding really stick in the mind. I’ve seen this over and over in my tutoring.

Stories from My Tutoring Work

The kind of story I’m talking about uses an extended-metaphor, and this way of teaching  is particularly helpful when you’re teaching algebra. Ask yourself: what would you rather have? Students scratching their heads (or tearing out their hair) to grasp a process taught as a collection of abstract steps? Or students grasping  a story and quickly seeing how it guides them in doing the math? I think the answer is probably pretty clear. So with this benefit in mind, let’s explore another story that teaches a critical algebraic skill: the skill of  “unpacking” terms locked inside parentheses.

To get the picture, first imagine that each set of parentheses, weirdly or not, represents a corrugated cardboard box, the kind that moving companies use to pack up your possessions. Extending this concept, the terms inside parentheses represent the items you pack when you move your goodies from one house to another.  Finally, for every set of parentheses (the box), imagine that you’ve hired either a good moving company or a bad moving company. (You can use a good company for one box and a bad company for a different “box” — it changes.) How can you tell whether the moving company is good or bad? Just look at the sign to the left of the parentheses. If the moving company is GOOD, you’ll see a positive sign to the left of the parentheses. If the moving company is BAD, you’ll spot a negative sign there.

Here’s how this idea looks:

+ (    )     The + sign here means you’ve hired a GOOD moving company for this box of stuff.

– (    )     This – sign means that you’ve hired a BAD moving company to pack up this box of things.

Now let’s put a few “possessions” inside the boxes.

+ (2x – 4)  This means a GOOD moving company has packed up your treasured items: the 2x and the – 4.

– (2x – 4)  Au contraire! This means that a BAD moving company has packed up the 2x and the – 4.

[Remember, of course, that the term 2x is actually a + 2x. No sign visible means there’s an invisible + sign before the term.]

What difference does it make if the moving company is GOOD or BAD? A big difference! If it’s a GOOD company, it packs your things up WELL.  Result: when you unpack your items, they come out exactly the same way in which they went into the box. So since a good moving company packed up your things in the expression:  + (2x – 4), when you go to unpack your things, everything will come out exactly as it went in. Here’s a representation of this unpacking process:

+ (2x – 4)

=      + 2x – 4

Note that when we take terms out of parentheses, we call this “unpacking” the terms. This works because algebra teachers fairly often describe the process of taking terms out of (   ) as “unpacking” the terms. So here’s a story whose rhetoric  matches the rhetoric of the algebraic process. Convenient, is it not?

Now let’s take a look at the opposite situation — what happens when you work with a BAD (boo, hiss!) moving company. In this case, the company does such a bad job that when you unpack your items, each and every item comes out  “broken.” In math, we indicate that terms are “broken” by showing that when they come out of the (  ), their signs,  + or – signs, are the EXACT OPPOSITE of what they should be. So if a term was packed up as a + term, it would come out as a – term.  Vice-versa, if it was packed up as a – term, it would come out as a + term. We show the process of unpacking terms packed by a BAD moving company, as follows:

– (2x – 4)

=      – 2x + 4

And that pretty much sums up the entire process. Understanding this story, students will be able to “unpack” terms from parentheses, over and over, with accuracy and understanding.

But since Practice Makes Perfect, here are a few problems to help your kiddos perfect this skill.

PROBLEMS:

“Unpack” these terms by removing the parentheses and writing the terms’ signs correctly:

a)  – (5a + 3)

b)  + (5a – 3)

c)  – (– 3a + 2b – 7)

d)  + (– 3a + 2b – 7)

e)  6 + (3a – 2)

f)  6 – (3a – 2)

g)  4a + 6 + (– 9a – 5)

h)  4a + 6 – (– 9a – 5)

ANSWERS:

a)  – (5a + 3)   =   – 5a – 3

b)  + (5a – 3)  =  + 5a – 3

c)  – (– 3a + 2b – 7)  =  + 3a – 2b + 7

d)  + (– 3a + 2b – 7) = – 3a + 2b – 7

e)  6 + (3a – 2)  =  + 3a + 4

f)  6 – (3a – 2)  =  – 3a + 8

g)  4a + 6 + (– 9a – 5)  =  – 5a + 1

h)  4a + 6 – (– 9a – 5)  =  + 13a + 11


Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like how Josh explains these problems, you’ll certainly  like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com  Just click the links in the sidebar for more information! 

How to Combine Positive & Negative Numbers — Quickly and Easily


If you or someone you know struggles when combining numbers with opposite signs — one positive, the other negative — this post is for you!

To be clear, I’m referring to problems like these:

 – 2 + 7 [first number negative, second number positive], or

+ 13 – 20 [first number positive, second number negative]

To work out the answers, turn each problem into a math-story. In this case, turn it into the story of a tug-of-war battle. Here’s how.

In the first problem, – 2 + 7, view the – 2 as meaning there are 2 people on the “negative” team; similarly, view the + 7 as meaning there are 7 people on the “positive” team.

There are just three things to keep in mind for this math-story:

1)  Every “person” participating in the tug-of-war is equally strong.

2)  The team with more people always wins; the team with fewer people always loses.

3)  In the story we figure out by how many people the winning team “outnumbers” the other team. That’s simple; it just means how many more people are on that team than are on the other team. Example: if the negative team has 2 people and the positive team has 7 people, we say the positive team “outnumbers” the negative team by 5 people, since 7 is 5 more than 2.

Now to simplify such a problem, just answer three simple questions: 

1)  How many people are on each team?
In our first problem, – 2 + 7, there are 2 people on the negative team and 7 people on the positive team.

2)  Which team WINS?
Since there are more people on the positive team, the positive team wins.

3) By how many people does the winning team OUTNUMBER the losing team?
Since the positives have 7 while the negatives have only 2, the positives outnumber the negatives by 5.

Now ignore the answer to the intro question, Question 1, but put together your answers to Questions 2 and 3.

ANSWER TO QUESTION 2:  +

ANSWER TO QUESTION 3:  5

ANSWERS TOGETHER:  + 5

All in all, this tells us that:  – 2 + 7 = + 5

For those of you who’ve torn your hair out over such problems, I have good news …

… THEY REALLY ARE THIS SIMPLE!

But to believe this, it will help to work out one more problem:  + 13 – 20.

Here, again, are the common-sense questions, along with their answers.

1)  How many people are on each team?
In this problem, + 13 – 20, there are 13 people on the positive team and 20 people on the negative team.

2)  Which team WINS?
Since there are more people on the negative team in this problem, the negative team wins.

3) By how many people does the winning team OUTNUMBER the losing team?
Since the negatives have 20 while the positives have only 13, the negatives outnumber the positives by 7.

Just as you did in the first problem, put together your answers to Questions 2 and 3.

ANSWER TO QUESTION 2:  

ANSWER TO QUESTION 3:  7

ANSWERS TOGETHER:  – 7

All in all, this tells us that:  + 13 – 20  = – 7

Now try these for practice:

a)  – 3 + 9

b) + 1 – 4

c)  –  9 + 23

d)  – 37 + 19

e) + 49 – 82

Answer to Practice Problems:

a)  – 3 + 9 = + 6

b) + 1 – 4 = – 3

c)  –  9 + 23 = + 14

d)  – 37 + 19 = – 18

e) + 49 – 82 = – 33

Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like the way Josh explains these problems, you will very likely like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com  Just click the links in the sidebar for more information! 

How to Decrease Algebraic Mistakes – Part 5


This is the fifth in a series of posts on how to help students make fewer mistakes in algebra.

No Mistakes

Let's Reduce Mistakes in Algebra!

So far I have introduced a form of notation I have developed, the double-slash, which looks like this:

//

and I have described some of the ways that students can use it.

I’ll continue the conversation by showing how this notation can help students combine like terms with greater care.

(more…)

Fun Math Problem #2


Here is the second in my series of “Fun Math Problems.”

Feel free to try these problems. Share them with friends and colleagues. Use them however you see fit! I will post the answer to the problems two days later, after people have had time to respond.

To provide your response, simply send an email to me @ info@SingingTurtle.com
and make your Subject: Fun Problem.
Please show how you worked the problem. Thanks. I will post the names of the first three people who get this right.

The Problem:  Before you go out to lunch, you glance at the clock above your desk. When you come back from lunch, you glance at the clock again, and you notice something strange. The minute and the hour hand have exchanged places from the positions they had just before you went to lunch.

The question is:  how long were you away?

Rusting clock face

Image by The Hidaway (Simon) via Flickr

Answer, Fun Math Problem


Answer to problem about the circular and square pegs and holes.

The “fit” for each situation is the following ratio:
(Area of Inner Figure) ÷ (Area of Outer Figure)

For the square peg in a round hole —
Call the radius of the circle r.
Then the diagonal of square “peg” = 2r
Notice that by slicing the square along its diagonal,
we get a 45-45-90 triangle, with the diagonal being
the hypotenuse and the sides being the two equal legs.
Using the proportions in a 45-45-90 triangle,
side of square peg = r times the square root of 2
Multiplying this side of the square by itself gives
us the area of the square, which comes out as:
2 times the radius squared

This being the case,
Area of square is: 2 times radius squared, and
Area of circle is: Pi times radius squared, and so …

Cancelling the value of the radius squared, we get:
Ratio of (Area of square) to (Area of circle) is:
2÷Pi = 0.6366

For the round peg in a square hole —
Call radius of the circle r.
And since the diameter of the circle is the same length as
the side of the square, the side of the square = 2r
Multiplying the side of the square by itself to get the
area of the square, we find that the area of the square
is given by: 4 times radius squared.

This being the case,
Area of circle is: Pi times radius squared
Area of square is: 4 times radius squared, and so …

Ratio of (Area of circle) to (Area of square) is therefore:
Pi ÷ 4 = 0.7854

Of the two ratios, the ratio of the circular peg in a square hole
is greater than that of the square peg in a circular hole.

Therefore we can say that the circular peg in a square hole
provides a better fit than a square peg in a circular hole.

And that is the answer!