Denesting Radicals

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(C)Copyright 2020, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

It's not easy getting them to leave the nest!

I was prepared to explain how to "denest" a nested radical, such as SQRT(3 + SQRT(5)), but aside from the obvious limitations of text on this blogging platform, I reviewed not only the steps, but the necessary conditions for denesting.

It's a little more involved than I usually get. So allow me to point to a wiki page on the subject.

As for the example above (which is hinted at in the comic), SQRT(3 + SQRT(5)) is equivalent to (1 + SQRT(5)) / SQRT(2), or SQRT(2)/2 + SQRT(5/2). You can check my math at wolframalpha.

I realize the imagery (and dialogue) is similar to my Free Radical comic, but what are you going to do? Math always circles back on itself.

For something else that is (sadly) neither free nor radical (I don't think it is), I have a flash fiction anthology available: In A Flash 2020, which you might enjoy reading next time you leave your nest and ride public transit, or when settling in your nest for a quiet evening.

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Rationalize the Denominator, Part 2

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(C)Copyright 2019, C. Burke.

At least he gave them a common denominator to start with.

Long-time readers will recognize the brown-haired girl as Bibi, from previous comics. I've been referring to the other two (in a non-canonically way) as Freedom and Serenity. Why I'm doing that is left as an exercise for the reader.




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Rationalize the Denominator

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(C)Copyright 2019, C. Burke.

Who says the denominator has to be rational anyway? Other than the curriculum? And my old teachers? And the text books?




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(x, why?) Mini: Radical

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(C)Copyright 2017, C. Burke.

Just watch out for those antioxidants




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Radical

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(C)Copyright 2017, C. Burke.

He's definitely lower than 3, but he's still more than 2.




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Irrational Fractions

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(C)Copyright 2016, C. Burke.

He's irrational, two.

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Multiplying Conjugates

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(C)Copyright 2016, C. Burke.

It's sort of the Reverse Lazarus Effect.

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Daily Regents: Graphing a Square Root Function

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams begin next month. At least, that is the plan.

June 2014, Question 25

Draw the graph of y = SQRT(x) - 1 on the set of axes below.
As show in the picture, SQRT means "square root", but I can't type that. However, keep that function in mind when we get to the calculator.

Here is what you saw on the paper:

On your graphing calculator, press the "Y=" key. Look at Figure 1 below.
Press "2nd" and "x²" to get the square root symbol and the open parenthesis.
Press the variable and then don't forget to close the parentheses! If you don't close them, then the "- 1" will be under the radical, and that is NOT what you want.
Hit GRAPH and you'll get Figure 2. This is what you need to graph on your paper.

The easiest way to get the values you need is to check the Table of Values. "2nd" and "Graph". Figure 3 and Figure 4 show you are of the coordinate pairs from 0 to 10. Notice that -1 and -2 are ERROR because they are out of the domain of the function. You won't have any line on the left side of the graph.

Plot the four points that have whole number values and draw a curve through them. Do not connect the four point with three straight lines.

Your final answer should look like this:

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Strange Square Roots

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(C)Copyright 2015, C. Burke.

Do NOT show my students this! They might think it's a rule and not a curiosity!

In fact, as strange as it looks, it isn't the only mixed number that has this property. And there isn't any reason to drop a jaw to the desk to figure out the pattern.

Let x be the whole number and y be the fraction (between 0 and 1, exclusive).

Then the equation says that (x)(sqrt(y)) = sqrt(x + y).
Square both sides we get (x²)(y) = x + y.
Subtract y from both sides (x²)(y) - y = x.
Factor the left side (y)(x² - 1) = x.
Finally, divide y = x/(x² - 1).

So you can pick any whole number value of x -- 2, 3, 10, whatever -- and substitute on the right side. You will get a value of y which is the fractional part of the mixed number that makes the "strange" square root work.

On a historical note: This is comic #1066. If you thought I'd do something about The Battle of Hastings ... well, it had crossed my mind, but too complicated and no way to plan in advance with the crazy schedule I'm keeping.




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Coming Soon...

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(C)Copyright 2015, C. Burke.

This was actually a leftover 'surd' joke, but I couldn't do much with 'To Surd, With Love'.




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Absurd

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(C)Copyright 2015, C. Burke.

The word "surd" is an absurd word.

For anyone who doesn't know what a "surd" is, I was one of you just a few weeks ago. I began brushing up on the Algebra 2 curriculum, in case I need to teach it in the fall, and I came across a page dedicated to the topic Surds.

Surds are just another name for radical numbers, particularly those that are irrational so the radical sign cannot be removed.

I checked with my son (the recent contributor and university student), and he hadn't heard the term. And then, a few days ago, a colleague in England, whom I follow on Twitter, posted some problems. The first response was "Surds! I love that topic!"

I admitted that I'd only recently heard that word, and the commenter asked, "Then what do you call them?"

The original poster responded, "Radicals!", so at least that term is known on the other side of the Pond.




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What's a Conjugate?

In Algebra, what is a conjugate? First, it's a noun, not a verb, and it's pronounced something like CON-juh-git, depending on your regional accent, but NOT as con-jyoo-GATE, like a big Language Arts scandal blasted across front pages of the tabloids.

The conjugate of a binomial, an algebraic expression with two terms, is a second binomial with the same terms but the sign between them has changed from plus to minus or minus to plus.

For example, 3x - 7 and 3x + 7 are conjugates.

What makes them interesting? One property of conjugates is to make things GO AWAY, and if there's one thing that Algebra students like is when things go away. And since I refuse to leave, this is the next best thing.

Add, Subtract, Multiply

If you add two conjugates, you double the first term and eliminate the second: (3x - 7) + (3x + 7) = 6x
If you subtract two conjugates, you elimated the first term and double the second: (3x - 7) - (3x + 7) = -14
-- keeping the sign of the term in the first binomial.

If you multiply them, something interesting happens:

You get a Difference of Squares. That is, the square of the first term minus the square of the second term. When you do the Distributive Property, you should get two more terms -- and don't you forget that! -- but in this case, those terms will cancel out! (-xy) + (xy) = 0.
(3x - 7)(3x + 7) = 9x² + 21x - 21x - 49 = 9x² - 49.

This can be useful not just for multiplying binomials, but for multiplying actual, honest-to-goodness, Real numbers, too!
Take, for example, (16) X (24). Not really easy to do in your head, but if you split the difference, you can see that it is the same as (20 - 4)(20 + 4).

Voila! The answer is 384! Wasn't that easy? Isn't this the greatest trick?

Nah, it's not. Just Kidding, really.

Just use a calculator. Seriously. No one really wants to square "bad" numbers in their head and then subtract them! But sometimes, it's kinda cool and you can impress your friends if you carefully pick your numbers!

Radicals!

BUT WAIT, THERE'S MORE!

Suppose you have a binomial where one of the terms is a radical number. Wouldn't you like that to go away, too? Well you can! Just multiply it by the conjugate.

I know, I know what you're thinking. You're thinking, "Yeah, that's okay, Mr. Burke. I'm cool with the radical. I'll just leave it alone!"

That's nice that you're cool with it, but you can't leave it alone. Suppose you have two divided by (6 plus radical 7). If there's a radical in the denominator of a fraction, it has to go away. That's just the rule. We're going to "simplify" it by multiplying both the numerator and the denominator of the fraction the conjugate, like this:

Isn't that so much better? It is, isn't it? Worth it, right? Right?

Imaginary Numbers

The same way that conjugates work for radical numbers, they can work with imaginary numbers.

If you have 3 + 4i, for example, in the bottom of a fraction again, you can make it real by multiplying by the conjugate, 3 - 4i.

Using our rule from about (3 - 4i)(3 + 4i) = 9 + 16 = 25, which looks suspiciously like a part of a Pythagorean Theorem problem -- but that's for another night.

Perimeter, Right Triangles and Radicals

In my Algebra class today, we were reviewing the process for adding and subtracting radical numbers. Previously, we simplified irrational numbers, such as the square of 80, which becomes 4 times the square root of 5.

To give them a more thoughtful question than just what is the sum of SQRT(75) + SQRT(48), where the only "thought" is to get past thinking that it's SQRT(123), I decided to pull out Right Triangles and that Old Favorite, the Pythagorean Theorem. Now, I didn't want them to just simplify the irrational number, I wanted some kind of addition in the problem. That brings us to Perimeter of a Right Triangle.

There are basically two types of problems you can offer up for consideration: problems with one radical number, and problems with two radical numbers. Three is just being mean, and overly complicates things -- on the other hand, it could make it interesting. Give the students the length of the legs (the base and the height), establish that there is a right angle between them, and let them go to work.

In the first kind of problem, you can pick any two numbers and you'll most likely have an irrational hypotenuse. Okay, but boring. It's more interesting if you make one of the legs irrational in such a way to make the hypotenuse rational. Surprise them.

It keeps it interesting when you consider these two problems, which look very similar, but are very different.

Consider the square root of 13 triangle first. If we square 6, we get 36. If we take the square of the square root of 13, we get 13. 36 + 13 = 49, which is the square of the hypotenuse. Therefore, the hypotenuse is 7.

Finding the perimeter is as simple as adding the three sides, which in this case means combining the like terms, which would be the integers 6 and 7. The perimeter is 13 + root 13.

In the square root 12 problem, the numbers had to be carefully selected. In this case, there will be two irrational numbers. If anything is to be combined, then the radicals have to simplify to the same radicand. Otherwise, the exercise is pointless.

If we square 6, we get 36 again. If we take the square of the square root of 12, we get 12, of course. 36 + 12 = 48, which is the square of the hypotenuse. Therefore, the hypotenuse is root 48.

We can't add any of the numbers as they are written, but we can simplify both of the radicals, as shown:

Both of the radicals have the square root of 3 in simplest form, and their coefficients can be added. The perimeter is 6 + 6 square root 3.

Interestingly, in both examples I wrote for lessons today, a double number appeared in the answer. This is actually something else to be careful about. Some students might see that as a co-incidence. Others might see it as a pattern and come to expect it.

There is an easy solution to that: give them some more problems to work on!

Math Can Be Painful

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(C)Copyright 2014, C. Burke.

The puns can be deadly.

That's my attempt at constructing the Gowanus Canal. It's a trendy area now, even though it still stinks. (Literally, the smell is nauseating.) However, the boats aren't really as big as they are on the map. Nor do they float on their sides, as seen in this aerial view.

N-RN.2 (Real Number System) - Rationalizing the Denominator

This is the third and final column on Common Core Standard N-RN.2. If you missed the first two parts, Part 1 dealt with evaluating expressions with rational exponents, and Part 2 showed how to simplify using factor trees and how to add and subtract radicals. The last piece of this standard (and since I'm only dealing with part ".2", I could really call it a "substandard" if I wanted to, mockingly) is to "Simplify radical expressions by rationalizing the denominator (Algebra 1 - EE.2)".

Previously, we mentioned that you can multiply two radical numbers by multiplying their radicands. We also factored radical numbers in order to simplify them. Let's talk about division. When you divide, you multiply by the reciprocal; that is, you can create a fraction of the two numbers without relying on your early education "gazintas". (You remember, "2 gazinta 6 three times".) Likewise, when you take the square root of a fraction, you are actually dividing one radical number by another.

So if you wanted the square root of 1/4, you would take the square root of the numerator (radical 1 is 1) over the square root of the denominator (radical 4 is 2). The result would be 1/2.

But suppose we wanted the square root of 1/2? Again, we can split it up into the square root of 1 (which is 1) over the square root of 2.

Here's where we run into a problem because there are rules from fractions. One of them is that there cannot be any radicals in denominator. You have to get rid of them.

We haven't discussed this before, but there's really only one simply way to get rid of a square root sign: square the number. We need to multiply the denominator by radical 2. We are allowed to do this because it's a fraction and we won't change the value of the fraction at all as long as we multiply the numerator by the same amount as the denominator. The fractions (square root of 2 over square root of 2 looks scary to evaluate until you remember that any number, even an irrational, divided by itself is one, with the exception of zero. If you multiply a fraction by 1, it doesn't change its value, even if it looks different. The result is that the radical is gone from the denominator and has moved into the numerator, which is allowed.

One more example. Try it yourself before scrolling down and looking at the image. What is the square root of 4/5?
Take the square root of each number. Rationalize the denominator. What's left in the numerator? What's left in the denominator?

Okay, check your work.

That's it for this standard. Time to move on to part 3, coming soon.

N-RN.2 (Real Number System) - Dealing With Radicals

In a recent post, we explored evaluating expressions with rational exponents in them, but there's more to Common Core Standard N-RN.2. Don't worry, some of it's easier to deal with than what we already tackled.

These are the items listed below the standard, at least according to the IXL website, where I found the list:

• Simplify radical expressions (Algebra 1 - EE.1)
• Simplify radical expressions by rationalizing the denominator (Algebra 1 - EE.2)
• Multiply radical expressions (Algebra 1 - EE.3)
• Add and subtract radical expressions (Algebra 1 - EE.4)
• Simplify radical expressions using the distributive property (Algebra 1 - EE.5)

Is there anything else to deal with? I don't know. Dealing with rational exponents isn't in this list, and yet I think that they might be encountered before Algebra 2. FYI, the notation "EE" stands for Expressions and Equations, which allows me to once again state that "Expressions don't have equal signs and are evaluated, and Equations do have equal signs and are solved."

Simplifying radical expressions is not a difficult task -- as long as you know that it does NOT mean pushing buttons on your calculator and coming up with an approximate decimal equivalent to 8 or 12 or 15 decimal places. Simplifying a radical is similar to reducing a fraction to its lowest terms. It makes it easier to deal with for computations (particularly adding and subtracting, when the radicals have to be "like terms") and comparisons. If the only thing you're planning to do with a radical number is square it, then, yes, simplifying it is a bigger waste of time than converting an improper fraction into a mixed number when it's only going to be used for slope.

There is a very straightforward method of simplifying square roots, but it seems to mystify some of my students who, apparently, never grasped the concept of what a square root (or a perfect square) was in the first place. They memorize steps, but uncertainty about the order causes them to mess up at the very end, removing radical signs from irrational numbers or leaving them in after taking a square root. (For example, they'll write that the square root of nine = the square root of three, instead of three.)

The simplest method involves finding the largest perfect square which is a factor of the radicand (i.e., the number under the radical sign). If it isn't the largest perfect square, then the radical hasn't been fully simplified. An example:

One problem my students face is not understanding the concept of a perfect square, so instead of 25 and 2, then use 5 and 10. After that, they're stuck, or they just decide, for example, that the square root of 5 is the same as 5 without the radical sign.

Because of this, I tried a different approach, using factor trees. They remembered doing them in middle school, and actually liked using them again. (You see, your teacher was right! You are using them again!) The example looked something like this instead:

After they have the prime factorization under the radical, I have them circle the pairs of numbers, cross them out and write one factor outside the radical. This has two downsides to it: first, if the number has a lot of factors, there will be a lot of extra work (but at least they will know, for certain, that they simplified their answer); second, if they don't complete the problem, they basically just drew a factor tree, which looks kinds childish and silly from a high school student.

Multiplying, Adding and Subtracting Radicals

Multiplying two radicals is as simple as multiplying two fractions. Just multiply the numbers under the radicand. For instance, radical 7 times radical 10 equals radical 70. If the number can be simplified, do it, according to the rules above. Obviously, if you square a radical, such as radical 6 times radical 6, the radical symbol goes away. In this case, you get radical 36, which is just 6.

As mentioned above, if you want to add or subtract radicals, they have to be alike. You can't add or subtract the following the way they are:

They aren't alike. It's like two to add 5² + 4² and getting 9². (In other words, you don't.)

But if you simplify the radicals, how to combine them becomes much clearer:

Finally, there is Division, but I'll save that for another column because of the standard, above, Simplify radical expressions by rationalizing the denominator.

Putting Down Roots

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(C)Copyright 2014, C. Burke.

They're always trying to put down those radicals!

I haven't hand-drawn one in a while. This one was touched up on the PC, instead of going over the original with a Sharpie.

Algebra 2/Trigonometry Regents for the Algebra 1 Student & Teacher

Today was the New York State Algebra 2/Trigonometry Regents exam. I don't teach this course, so I won't comment personally on how good a test it was for Trig students, other than to say that a couple of colleagues called it a "fair exam". What I can say about this exam is this: Algebra 1 teachers can use many of these multiple choice questions in their own classes with little to no adjustments. If I might so boldly and "arrogantly" claim, the top students in my Algebra 1 class could have solved 8 of the first 9 problems. An above average student would've gotten at least five of those correct.

With this in mind, I'd like to once again go over the Algebra 2 problems which I believe Algebra 1 students could handle, even if only as challenge problems.

Algebra 2/Trigonometry

1. Which survey is least likely to contain bias?
1. surveying a sample of people leaving a movie theater to determine which flavor of ice cream is the most popular
2. surveying the members of a football team to determine the most-watched TV sport
3. surveying a sample of people leaving a library to determine the average number of books a person reads in a year
4. surveying a sample of people leaving a gym to determine the average number of hours a person exercises per week

Not having my students the entire year, I didn't get to "bias" in Common Core Algebra (I believe the previous teacher should have touched on it). I know it was covered in the Integrated Algebra course. The second, third and fourth choices are going to places to ask a question pertaining to the place where the questions are asked; e.g., readers at a library. Only the first one goes to a place where you will find different types of people, not just ice cream lovers. Could there be bias in Choice 1? Of course, it could. Not all people go to movies. But it is still less biased than the other three.

2. The expression (2a)^-4 is equivalent to ...?

If you know that a negative exponent means to (basically) take the reciprocal, then you'll get 1/(16a⁴ as your answer.

Question 3 is a trigonometry question. We'll skip that.

4. Expressed in its simplest form,

is

This could easily be used in Algebra 1 without the negatives under the radicals. It could be used as an extension if there's time. Some of my students knew about imaginary numbers, even if they weren't sure exactly what they were. And they knew they had something to do with square roots.

It's also easy to reason out the answer from the choices. Once you realize that i is involved in both radicals and can be factored out, you've eliminated choices (1) and (2). Realizing that you're subtracting a bigger number from a smaller number indicates that the answer will be negative, eliminating choice (4). (3) is the answer.

5. Theresa is oomparing the graphs of y = 2^x and y = 5^x. Which statement is true?

First of all, both graphs have a y-intercept of (0, 1). Choices (1) and (4) are silly. (Really, "neither graph has a y-intercept"?) Of the two, y = 5^x is steeper. You can check this in your graphing calculator if you weren't sure.

6. The solution set of the equation

is

For Algebra 1 students (and some Trig students), the fastest method is to plug in the choices. Trying -2 doesn't work. Trying 2 does work. Only one solution set contains 2. It also contains 4, which also works.

How are you supposed to solve this? Square both sides and solve the resulting quadratic equation. For multiple choice, plugging in is much faster.

7. The expression is equivalent to

(2)(2) = 4; (-3)(x)^.5 X (-3)(x)^.5 = 9x; (2)(2)(-3)(x)^.5 = -12(x)^.5
The correct choice is (3).

8. Which step can be used when solving x² - 6x - 25 = 0 by completing the square.

Okay, I never did completing the square in Integrated Algebra. It might've been there in the textbook, but it wasn't covered in the curriculum, and it wasn't on the Algebra Regents. That said, it was in the Common Core Algebra this year, and my students picked it up pretty easily. (Well, most of them did.)

To complete the square, you need to halve the -6, getting -3, and then squaring that, getting 9. So +9 is added to each side of the equation and +25 is also added to each side of the equation to get rid of the -25 on the left. The correct choice is (1).

9. Which graph represents a function?

Seriously? This is an Algebra 1 question. If there aren't two y values for the same x-value, then it is a function. Choice (1).

Question 10 is a trigonometry question. We'll skip that.

11. What is the common difference of the arithmetic sequence below?
-7x, -4x, -x, 2x, 5x, . . .

Algebra students should recognize the pattern and deduce that the "common difference" is 3x.

Jumping ahead...

14. What is the product of the roots of the quadratic equation 2x² - 7x = 5?

I should include questions like this. There's no reason not to, and it will get an extra step of them. First solve the quadratic equation, and then multiply the roots. The only problem I have with this -- and maybe it isn't a problem at all -- is that the most common mistake my students make in solving quadratics in flipping the sign. If they flipped both signs and then multiply the answer, then the mistakes will cancel out.

Quick use of the quadratic formula will get you ... two radical conjugates. Okay, so this goes beyond the scope of Integrated Algebra, but a teacher could modify this one a little. But anyway, the product is one-sixteenth of (49 - 89), which is -5/2.

It's actually simpler than this: the rule for product of roots is c/a, which is -5/2. Introducing this right after doing a long problem might be a good way to make them remember the shortcut. It also reinforces the fact that if you can't remember the formulas and shortcuts, it helps to know where they come from, so you can derive them if you have to.

* * *

Continuing the thread...

15. What is the equation of the circle passing through the point (6, 5) and centered at (3, -4)?

This question gets asked on the Geometry Regents at least 3 or 4 times on every test. The only difference here is that the radius is an irrational number, but big deal. Geometry students need to deal with irrational numbers, and the square of the number is needed anyway. (6 - 3)² + (5 - -4)² = 90. So the equation is
(x - 3)² + (y + 4)² = 90.

16. The formula to determine continuously compounded interest is A = Pe^rt, where A is the amount of money in the account, P is the initial investment, r is the interest rate and t is the time, in years. Which equation could be used to determine the value of an account with an $18,000 initial investment, at an interest rate of 1.25% for 24 months?

As complicated as this looks, this is a simple substitution question. It could be given to my freshmen as an extension, just to see if they really can parse a question. You don't have to explain e yet, if you don't want to be, because it could be considered just any variable for the moment. (I realize that it's a constant, but let's not confuse matters at the moment.) The only "trick" to the problem is to remember that 24 months is 2 years. This trips up some students with I=PRT, too.

Question 17 is interesting. Without the "+ 1", it's a simple proportion that leads to a quadratic equation if you don't factor the difference of squares and multiply the fraction on the right by (x + 3)/(x + 3). The "+ 1" makes the addition a little more interesting. Lots of possibilities with this equation for Algebra students.

18. The graph below shows the average price of gasoline, in dollars, for the years 1997 to 2007. [GRAPH NOT SHOWN] What is the approximate range of this graph?

Seriously? Range measures the y values on the graph. The lowest point appears to be about 1.00 or lower, and the highest point is between 2.00 and 2.50. The correct choice would be 0.97 < y < 2.38. Choices 1 and 2 relate to the domain of the graph.

What are your opinions of all this?