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Difficulty Level: At Grade Created by: CK-12

Name: __________________

As a student of mine once said, “In real life, no one ever says, ‘Here’s \begin{align*}100\;\mathrm{dollars}\end{align*}, what’s the square root of it?’” She’s right, of course—as far as I know, no one ever takes the square root of money. And she is asking exactly the right question, which is: why do we need roots anyway?

1. If a square is \begin{align*}49\;\mathrm{ft}^2\end{align*} in area, how long are the sides?

Q. You call that real life?

A. OK, you asked for it...

2. A real estate developer is putting houses down on a plot of land that is \begin{align*}50\end{align*} acres large. He wants to put down \begin{align*}100\end{align*} houses, so each house will sit on a \begin{align*}\frac{1}{2}\end{align*}-acre lot. (\begin{align*}1\end{align*} acre is \begin{align*}43,560\;\mathrm{square \ feet}\end{align*}.) If each house sits on a square lot, how long are the sides of each lot?

3. A piano is dropped from a building \begin{align*}100\;\mathrm{ft}\end{align*} high. (“Dropped” implies that someone just let go of it, instead of throwing it—so it has no initial velocity.)

a. Write the equation of motion for this piano, recalling as always that

\begin{align*}h(t)=h_o+v_ot-16t^2.\end{align*}

b. According to the equation, how high is the piano when \begin{align*}t=0\end{align*}? Explain in words what this answer means.

c. After \begin{align*}2\;\mathrm{seconds}\end{align*}, how high is the piano?

d. How many seconds does it take the piano to reach the ground?

e. Find the inverse function that will enable to me find the time \begin{align*}t\end{align*} when the piano reaches any given height \begin{align*}h\end{align*}.

Convinced? Square roots come up all the time in real life, because squaring things comes up all the time in real life, and the square root is how you get back. So we’re going to have a unit on square roots.

If I tell you that \begin{align*}\sqrt{25}=5\end{align*}, that is the same thing as telling you that \begin{align*}5^2=25\end{align*}. Based on that kind of logic, rewrite the following radical equations \begin{align*}(4-6)\end{align*} as exponent equations.

4. \begin{align*}\sqrt{100} = 10\end{align*}

5. \begin{align*}\sqrt[3]{8} = 2\end{align*}

6. \begin{align*}\sqrt[a]{b} = c\end{align*}

7. Now, rewrite all three as logarithm equations. (\begin{align*}*\end{align*}You mean we still have to know that?)

Some Very Important Generalizations

8. \begin{align*}\sqrt{9} = \end{align*}

9. \begin{align*}\sqrt{4} = \end{align*}

10. \begin{align*}\sqrt{9} \times \sqrt{4} = \end{align*}

11. \begin{align*}\sqrt{9 \times 4} =\end{align*}

12. Based on 8-11, write an algebraic generalization.

13. Now, give me a completely different example of that same generalization: four different statements, like 8-11, that could be used to generate that same generalization.

14. \begin{align*}\frac{\sqrt{9}} {\sqrt{4}} = \end{align*}

15. \begin{align*}\sqrt{\frac{9}{4}}\end{align*}

a.What do you think is the answer?

b.Test by squaring back. (To square anything, multiply it by itself. So this just requires multiplying fractions!) If it doesn’t work, try something else, until you are convinced that you have a good \begin{align*}\sqrt{\frac{9}{4}}\end{align*}.

16. Based on numbers 14-15, write an algebraic generalization.

17. \begin{align*}\sqrt{9} + \sqrt{4} = \end{align*}

18. \begin{align*}\sqrt{9 + 4} = \end{align*}

19. Based on numbers 17-18, write an algebraic generalization that is not true.

Based on the generalization you wrote in \begin{align*}^{\#}12\end{align*}, and given the fact that \begin{align*}4 \times 2 = 8\end{align*}, we can simplify \begin{align*}\sqrt{8}\end{align*} as follows.

\begin{align*}\sqrt{8} = \sqrt{4 \times 2}\end{align*} (because \begin{align*}8\end{align*} is the same thing as \begin{align*}4 \times 2\end{align*})

\begin{align*}\sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2}\end{align*} (this is where we use that generalization)

\begin{align*}\sqrt{4} \times \sqrt{2} = 2 \sqrt{2}\end{align*} because \begin{align*}\sqrt{4}\end{align*} is the same thing as \begin{align*}2\end{align*})

So, we see that \begin{align*}\sqrt{8}\end{align*} is the same thing as \begin{align*}2\sqrt{2}\end{align*}.

20. Find \begin{align*}\sqrt{8}\end{align*} on your calculator.

21. Find \begin{align*}\sqrt{2}\end{align*} on your calculator.

22. Double your answer to \begin{align*}^{\#}21\end{align*} on your calculator.

23. So, did it work?

24. Good! Then let’s try another one. Simplify \begin{align*}\sqrt{72}\end{align*}, using the same steps I took to simplify \begin{align*}\sqrt{8}\end{align*}.

25. Check your answer on the calculator. Did it work?

26. Oh yeah, one final question: what is \begin{align*}\sqrt{x^{16}}\end{align*}? How can you test your answer?

Name: __________________

For the following problems, I am not looking for big long unmanageable decimals—just simplified expressions. You should not have to use your calculator at all, except possibly to check your answers. Remember, you can always check your answer by squaring back!

1. \begin{align*}\sqrt{64}\end{align*}

2. \begin{align*}\sqrt[3]{64}\end{align*}

3. \begin{align*}3\sqrt{64}\end{align*}

4. \begin{align*}-\sqrt{64}\end{align*}

5. \begin{align*}\sqrt{-64}\end{align*}

6. \begin{align*}\sqrt[3]{-64}\end{align*}

7. If \begin{align*}\sqrt[n]{-x}\end{align*} has a real answer, what can you say about \begin{align*}n\end{align*}?

8. \begin{align*}\sqrt{8}\end{align*}

9. \begin{align*}\sqrt{18}\end{align*}

10. \begin{align*}\sqrt{48}\end{align*}

11. \begin{align*}\sqrt{70}\end{align*}

12. \begin{align*}\sqrt{72}\end{align*}

13. \begin{align*}\sqrt{100}\end{align*}

14. \begin{align*}\sqrt{\frac{1}{4}}\end{align*}

15. \begin{align*}\sqrt{\frac{4}{9}}\end{align*}

16. \begin{align*}\sqrt{\frac{8}{12}}\end{align*}

17. \begin{align*}\sqrt{16x^2}\end{align*}

a. Simplify as much as possible (just like all the other problems)

b. Check your answer with \begin{align*}x = 3\end{align*}. Did it work?

18. \begin{align*}\sqrt{(16x)^2}\end{align*}

19. \begin{align*}\sqrt{x^{10}}\end{align*}

20. \begin{align*}\sqrt{x^{11}}\end{align*}

21. \begin{align*}\sqrt{75x^3y^6z^5}\end{align*}

22. \begin{align*}\sqrt{x^2y^2}\end{align*}

23. \begin{align*}\sqrt{x^2 + y^2}\end{align*}

24. \begin{align*}\sqrt{x^2 + 2xy + y^2}\end{align*}

25. \begin{align*}\sqrt{x^2 + 9}\end{align*}

26. \begin{align*}\sqrt{x^2 + 6x + 9}\end{align*} Say, remember inverse functions?

27. \begin{align*}f(x)=x^2.\end{align*}

a. Find the inverse function.

b. Test it.

28. \begin{align*}f(x)=x^3.\end{align*}

a. Find the inverse function.

b. Test it.

29. \begin{align*}f(x)=3^x.\end{align*}

a. Find the inverse function.

b. Test it.

Name: _________________

A Bunch of Other Stuff About Radicals

Let’s start off with a bit of real life again, shall we?

1. Albert Einstein’s “Special Theory of Relativity” tells us that matter and energy are different forms of the same thing. (Previously, they were thought of as two completely different things.) If you have some matter, you can convert it to energy; if you have some energy, you can convert it to matter. This is expressed mathematically in the famous equation \begin{align*}E = mc^2\end{align*}, where \begin{align*}E\end{align*} is the amount of energy, \begin{align*}m\end{align*} is the amount of matter, and \begin{align*}c\end{align*} is the speed of light. So, suppose I did an experiment where I converted \begin{align*}m\end{align*} kilograms of matter, and wound up with \begin{align*}E\end{align*} Joules of energy. Give me the equation I could use that would help me figure out, from these two numbers, what the speed of light is.

2. The following figure is an Aerobie, or a washer, or whatever you want to call it—it’s the shaded area, a ring with inner thickness \begin{align*}r_1\end{align*} and outer thickness \begin{align*}r_2\end{align*}.

a. What is the area of this shaded region, in terms of \begin{align*}r_1\end{align*} and \begin{align*}r_2\end{align*}?

b. Suppose I told you that the area of the shaded region is \begin{align*}32 \pi\end{align*}, and that the inner radius \begin{align*}r_1\end{align*} is \begin{align*}7\end{align*}. What is the outer radius \begin{align*}r_2\end{align*}?

c. Suppose I told you that the area of the shaded region is \begin{align*}A\end{align*}, and that the outer radius is \begin{align*}r_2\end{align*}. Find a formula for the inner radius.

OK, that’s enough about real life. Let’s try simplifying a few expressions, using the rules we developed yesterday.

3. \begin{align*}\sqrt{100y}\end{align*}

4. \begin{align*}\sqrt{\frac{x}{25}}\end{align*}

5. \begin{align*}\sqrt{x + 16}\end{align*}

6. \begin{align*}\frac{\sqrt{50}}{2} = \end{align*}

7. \begin{align*}5\sqrt{2} + 2 \sqrt{3} - 3 \sqrt{2} = \end{align*}

8. \begin{align*}\sqrt{27} - \sqrt{48} = \end{align*}

Let’s try some that are a bit trickier—sort of like rational expressions. Don’t forget to start by getting a common denominator!

9. \begin{align*}\frac{1}{\sqrt{2}} + \frac{\sqrt{2}} {2} = \end{align*}

a. Simplify. (Don’t use your calculator, it won’t help.)

b. Now, check your answer by plugging the original formula into your calculator. What do you get? Did it work?

10. \begin{align*}\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}=\end{align*}

a. Simplify. (Don’t use your calculator, it won’t help.)

b. Now, check your answer by plugging the original formula into your calculator. What do you get? Did it work?

11. \begin{align*}\frac{1}{\sqrt{3} + 1} - \frac{\sqrt{3}} {2} = \end{align*}

a. Simplify. (Don’t use your calculator, it won’t help.)

b. Now, check your answer by plugging the original formula into your calculator. What do you get? Did it work?

And now, the question you knew I would ask...

12. Graph \begin{align*}y = \sqrt{x}\end{align*}.

a. Plot a whole mess of points. (Choose \begin{align*}x-\end{align*}values that will give you pretty easy-to-graph \begin{align*}y-\end{align*}values!)

b. What is the domain? What is the range?

c. Draw the graph.

13. Graph \begin{align*}y = \sqrt{x} - 3\end{align*} by shifting the previous graph.

a. Plug in a couple of points to make sure your “shift” was correct. Fix it if it wasn’t.

b. What is the domain? What is the range?

14. Graph \begin{align*}y = \sqrt{x - 3}\end{align*} by shifting the previous graph.

a. Plug in a couple of points to make sure your “shift” was correct. Fix it if it wasn’t.

b. What is the domain? What is the range?

Name: _________________

Homework: A Bunch of Other Stuff About Radicals

1. Several hundred years before Einstein, Isaac Newton proposed a theory of gravity. According to Newton’s theory, any two bodies exert a force on each other, pulling them closer together. The force is given by the equation \begin{align*}F = G \frac{m_1m_2}{r^2}\end{align*}, where \begin{align*}F\end{align*} is the force of attraction, \begin{align*}G\end{align*} is a constant, \begin{align*}m_1\end{align*} and \begin{align*}m_2\end{align*} are the masses of the two different bodies, and \begin{align*}r\end{align*} is the distance between them. Find a formula that would give you \begin{align*}r\end{align*} if you already knew \begin{align*}F, G, m_1,\end{align*} and \begin{align*}m_2\end{align*}.

2. In the following drawing, \begin{align*}m\end{align*} is the height (vertical height, straight up) of the mountain; \begin{align*}s\end{align*} is the length of the ski lift (the diagonal line); and \begin{align*}x\end{align*} is the horizontal distance from the bottom of the ski lift to the bottom of the mountain.

a. Label these three numbers on the diagram. Note that they make a right triangle.

b. Write the relationship between the three. (Pythagorean Theorem)

c. If you build the ski lift starting \begin{align*}1,200\;\mathrm{feet}\end{align*} from the bottom of the mountain, and the mountain is \begin{align*}800\;\mathrm{feet}\end{align*} high, how long is the ski lift?

d. If the ski lift is \begin{align*}s\end{align*} feet long, and you build it starting \begin{align*}x\;\mathrm{feet}\end{align*} from the bottom of the mountain, how high is the mountain?

3. Simplify \begin{align*}\frac{\sqrt{28}}{4 - \sqrt{7}}\end{align*}. Check your answer on your calculator.

4. Simplify \begin{align*}\frac{1}{\sqrt{5}} + \frac{1}{\sqrt{5} - 2}\end{align*}. Check your answer on your calculator.

5. Graph \begin{align*}y = \sqrt{x} - 3\end{align*}. What is the domain and range?

6. Graph \begin{align*}y = \sqrt{x - 3}\end{align*}. What is the domain and range?

Name: _________________

Before I get into the radical equations, there is something very important I have to get out of the way. Square these two out:

1. \begin{align*}(2 + \sqrt{2} )^2 =\end{align*}

2. \begin{align*}\left (\sqrt{3} + \sqrt{2}\right )^2 = \end{align*}

How’d it go? If you got six for the first answer and five for the second, stop! Go back and look again, because those answers are not right. (If you don’t believe me, try it on your calculator.) When you’ve got those correctly simplified (feel free to ask—or, again, check on your calculator) then go on.

Now, radical equations. Let’s start off with an easy radical equation.

3. \begin{align*}\sqrt{2}x + 3 = 7\end{align*}

I call this an “easy” radical equation because there is no \begin{align*}x\end{align*} under the square root. Sure, there’s a \begin{align*}\sqrt{2}\end{align*}, but that’s just a number. So you can solve it pretty much the same way you would solve \begin{align*}4x + 3 = 7\end{align*}; just subtract \begin{align*}3\end{align*}, then divide by \begin{align*}\sqrt{2}\end{align*}.

a. Solve for \begin{align*}x\end{align*}.

b. Check your answer by plugging it into the original equation. Does it work?

This next one is definitely trickier, but it is still in the category that I call “easy” because there is still no \begin{align*}x\end{align*} under the square root.

4. \begin{align*}\sqrt{2} x + 3x = 7\end{align*}

a. Solve for \begin{align*}x\end{align*}.

b. Check your answer by plugging it into the original equation. Does it work? (Feel free to use your calculator, but show me what you did and how it came out.)

Now, what if there is an \begin{align*}x\end{align*} under the square root? Let’s try a basic one like that.

5. Solve for \begin{align*}x\end{align*}: \begin{align*}\sqrt{x} = 9\end{align*}

What did you get? If you said the answer is three: shame, shame. The square root of \begin{align*}3\end{align*} isn’t \begin{align*}9\end{align*}, is it? Try again.

OK, that’s better. You probably guessed your way to the answer. But if you had to be systematic about it, you could say, “I got to the answer by squaring both sides.” The rule is: whenever there is an \begin{align*}x\end{align*} under a radical, you will have to square both sides. If there is no \begin{align*}x\end{align*} under the radical, don’t square both sides.

It worked out this time, but squaring both sides is fraught with peril. Here are a few examples.

6. \begin{align*}\sqrt{x} = -9\end{align*}

a. Solve for \begin{align*}x\end{align*}, by squaring both sides.

b. Check your answer by plugging it into the original equation.

Hey, what happened? When you square both sides, you get \begin{align*}x = 81\end{align*}, just like before. But this time, it’s the wrong answer: \begin{align*}\sqrt{81}\end{align*} is not \begin{align*}–9\end{align*}. The moral of the story is that when you square both sides, you can introduce false answers. So whenever you square both sides, you have to check your answers to see if they work. (We will see that rule come up again in some much less obvious places, so it’s a good idea to get it under your belt now: whenever you square both sides, you can introduce false answers!)

But that isn’t the only danger of squaring both sides. Check this out...

7. Solve for \begin{align*}x\end{align*} by squaring both sides: \begin{align*}2 + \sqrt{x} = 5\end{align*}

Hey, what happened there? When you square the left side, you got (I hope) \begin{align*}x + 4 \sqrt{x} + 4\end{align*}. Life isn’t any simpler, is it? So the lesson there is, you have to get the square root by itself before you can square both sides. Let’s come back to that problem.

8. \begin{align*}2 + \sqrt{x} = 5\end{align*}

a. Solve for \begin{align*}x\end{align*} by first getting the square root by itself, and then squaring both sides

b. Check your answer in the original equation.

Whew! Much better! Some of you may have never fallen into the trap—you may have just subtracted the two to begin with. But you will find you need the same technique for harder problems, such as this one:

9. \begin{align*}x - \sqrt{x} = 6\end{align*}

a. Solve for \begin{align*}x\end{align*} by first getting the square root by itself, and then squaring both sides, and then solving the resulting equation. (Note: you should end up with two answers.)

b. Check your answers in the original equation. (Note: if you did everything right, you should find that one answer works and the other doesn’t. Once again, we see that squaring both sides can introduce false answers!)

The more times you see \begin{align*}x\end{align*} under a square root, the more squaring you have to do. For instance...

10. \begin{align*}\sqrt{x - 2} = \sqrt{3} - \sqrt{x}\end{align*}

What do you do now? You’re going to have to square both sides...that will simplify the left, but the right will still be complicated. But if you look closely, you will see that you have changed an equation with \begin{align*}x\end{align*} under the square root twice, into an equation with \begin{align*}x\end{align*} under the square root once. So then, you can solve it the way you did above: get the square root by itself and square both sides. Before you are done, you will have squared both sides twice!

Solve \begin{align*}^{\#}10\end{align*} and check your answers...

Name: __________________

For each of the following, you will first identify it as one of three types of problem:

• No \begin{align*}x\end{align*} under a radical, so don’t square both sides.
• \begin{align*}x\end{align*} under a radical, so you will have to isolate it and square both sides.
• More than one \begin{align*}x\end{align*} under a radical, so you will have to isolate-and-square more than once!

Then you will solve it; and finally, you will check your answers (often on a calculator). Remember that if you squared both sides, you may get false answers even if you did the problem correctly! If you did not square both sides, a false answer means you must have made a mistake somewhere.

1. \begin{align*}\sqrt{x}= –3\end{align*}
1. Which type of problem is it?
2. Solve for \begin{align*}x\end{align*}.
2. \begin{align*}\sqrt{2}x - 3 = \sqrt{3}x\end{align*}
1. Which type of problem is it?
2. Solve for \begin{align*}x\end{align*}.
3. \begin{align*}x - \sqrt{2x} = 4\end{align*}
1. Which type of problem is it?
2. Solve for \begin{align*}x\end{align*}.
4. \begin{align*}\sqrt{4x + 2} - \sqrt{2x} = 1\end{align*}
1. Which type of problem is it?
2. Solve for \begin{align*}x\end{align*}.
5. \begin{align*}3 - \sqrt{x - 2} = -4\end{align*}
1. Which type of problem is it?
2. Solve for \begin{align*}x\end{align*}.
6. \begin{align*}x + 2 \sqrt{x} = 15\end{align*}
1. Which type of problem is it?
2. Solve for \begin{align*}x\end{align*}.
7. \begin{align*}\sqrt{x + 4} + \sqrt{x} = 2\end{align*}
1. Which type of problem is it?
2. Solve for \begin{align*}x\end{align*}.

Name: ________________________

1. Punch this into your calculator and give the answer rounded to three decimal places. This is the only question on the quiz where I want an answer in decimal form:

\begin{align*}\sqrt{69} = \end{align*}

2. Give me an approximate answer for \begin{align*}\sqrt[3]{10}\end{align*}

Simplify. Give answers using radicals, not decimals or approximations.

3. \begin{align*}\sqrt{400}\end{align*}

4. \begin{align*}\sqrt[3]{-27}\end{align*}

5. \begin{align*}3 \sqrt{-27}\end{align*}

6. \begin{align*}\sqrt{108}\end{align*}

7. \begin{align*}\sqrt{20} - \sqrt{45}\end{align*}

8. \begin{align*}\sqrt{\frac{300}{64}}\end{align*}

9. \begin{align*}\sqrt{x^{16}}\end{align*}

10. \begin{align*}\sqrt[5]{x^{38}}\end{align*}

11. \begin{align*}\sqrt{98x^{20}y^5z}\end{align*}

12. \begin{align*}\sqrt{4x^2 + 9y^4}\end{align*}

13. \begin{align*}\left (\sqrt{3} - \sqrt{2} \right )^2\end{align*}

14. \begin{align*}\frac{\sqrt{24}}{2 + \sqrt{2}}\end{align*}

15. \begin{align*}\sqrt[3]{\mathrm{something}} = x + 2\end{align*}. What is the something?

16. Rewrite as an exponent equation: \begin{align*}x = \sqrt[n]{y}\end{align*}

17. Rewrite as a radical equation: \begin{align*}a^b=c\end{align*}

18. Rewrite as a logarithm equation: \begin{align*}a^b=c\end{align*}

Solve for \begin{align*}x\end{align*}

19. \begin{align*}\frac{3 + \sqrt{x}} {2} = 5\end{align*}

20. \begin{align*}\sqrt{2x + 1} - \sqrt{x} = 1\end{align*}

21. \begin{align*}3x + \sqrt{3} (x) - 4 + \sqrt{8} = 0\end{align*}

22. \begin{align*}\sqrt{x} + \sqrt{2} (x) - \sqrt{2} = 0\end{align*}

23. For an object moving in a circle around the origin, whenever it is at the point \begin{align*}(x,y)\end{align*}, its distance to the center of the circle is given by: \begin{align*}r = \sqrt{x^2 + y^2}\end{align*}.

a. Solve this equation for \begin{align*}x\end{align*}.

b. If \begin{align*}y = 2\end{align*} and \begin{align*}r = 2 \frac{1}{2}\end{align*}, what is \begin{align*}x\end{align*}?

24. Graph \begin{align*}y = – \sqrt{x} + 3\end{align*}.

25. What are the domain and range of the graph you drew in \begin{align*}^{\#}24\end{align*}?

Extra credit: Draw a graph of \begin{align*}y = \sqrt[3]{x}\end{align*}.

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