Tier 3 Part 1 Exercises 2014 - Triad...

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Dr. Del’s Practical Math

Tier 3 Part 1

Problems and Answers

Lessons 1 - 9


A Message to the student regarding problems: As you know, the more you practice the better and faster you will get, just like in any game or sport. So these problems are for those of you who want to get better and better. If you solve the problems given in the Simmons book you will be OK and be able to pass the Quiz, and probably do pretty good on the SAT and ACT. But, the more you practice the better you’ll get. And, I find I enjoy math more the better I get. I have included some problems that should also help you understand the math behind them better and better. Some of these problems may be pretty challenging, although you do have enough knowledge to solve them if you think long enough. DO NOT get discouraged when you tackle a hard problem and struggle with it. THAT HAPPENS to all mathematicians. In fact, we really seem to be understanding math at a deeper level when we struggle with a problem. If you get frustrated, take a break. If you have others to work with help each other. You will learn a lot by teaching and explaining things to others. That is really how I learned math. I started “tutoring” some of my classmates when I was 15 years old. And, they helped me too. Learning math is like any sport, the more we practice and coach others the better we get. So, I hope you enjoy these additional problems. IF YOU FIND A MISTAKE in my answers, Bully for you. I’m pretty sure there will be some mistakes creeping in from time to time. Sometimes, I leave them in for you to find a little gem. It will help you build your confidence too.


Never believe something just because it is written in a textbook. Verify it with your own thinking and understanding as much as you can. I can’t tell you how many “mistakes” I have found in various textbooks over the years. And, sometimes on tests too. You might want to practice my Motto: “Never be Satisfied, Always be Content.” Above all, have fun and appreciate what we have been given and this opportunity to improve ourselves and open up other doors of understanding, which math does.


Lesson 1 Exercises Place each number in the smallest possible set N, Z, Q, R, C (Natural Numbers, Integers, Rational Numbers, Real Numbers, Complex Numbers) 1. 89 2. π/2 3. -13 4. 6/7 5. √49 6. √679 7. √729 8. √-75 9. 3+ π 10. -√81 11. √(49/25) 12. √ π 13. 1/√5 14. √2 + 2


Lesson 1 Exercise Answers Place each number in the smallest possible set N, Z, Q, R, C (Natural Numbers, Integers, Rational Numbers, Real Numbers, Complex Numbers) 1. 89 N Positive Integer 2. π/2 R It is irrational 3. -13 Z Negative Integer 4. 6/7 Q Rational Number 5. √49 N Positive Integer, 7 6. √679 R It is Irrational 7. √729 N It is positive integer, 27 8. √-75 C Complex, Not Real, Imaginary 9. 3+ π R It is Irrational 10. -√81 Z -9 11. √(49/25) Q Rational 7/5 12. √ π R It is Irrational 13. 1/√5 R It is Irrational 14. √2 + 2 R It is Irrational


Lesson 2 Exercises Simplify #1 through #3 and Factor #4 through #6 1. [(3x – 2y) – (5x + 4y) + 6y + 3x] 2. {2(a +2b +c) +(3a -6b + 2y) – 3(2b -3y)} 3. [1.2(4.5b -6.7a) – 2.3(1.9a + 2.7b)] 4. 12a - 18b + 24x 5. 4x2y3 – 6x3y2 6. 9a2bc2 – 12ab2c Combine and simplify: Note: (b – a) = -(a – b) (If you don’t understand exponents yet, wait until the next lesson for #7, 8, & 12.) 7. x2/y + y2/x 8. x/(x – 2) + x/(1 – x) Hint: Find common denominator 9. 1/[2 + 1/(x+3)] 10. 1/[1 + 1/(a-1)] Which problem in Simmons book is this equivalent to? 11. (a – b)/[1/a – 1/b] 12. (x – 1)(x3 + x2 + x + 1)


Lesson 2 Exercise Answers Simplify #1 through #3 and Factor #4 through #6 1. x 2. 5a -8b +2c + 11y 3. -12.41a -0.81b 4. 6(2a -3b +4x) 5. 2x2y2(2y – 3x) 6. 3abc(3ac – 4b) Combine and simplify: Note: (b – a) = -(a – b) (If you don’t understand exponents yet, wait until the next lesson for #7, 8, & 12.) 7. (x3 + y3)/xy 8. x/(x – 2) + x/(1 – x) = [x(1-x) + x(x-2)]/(x-2)(1-x) = -x/[(x-2)(1-x)] = x/[(x-2)(x-1)] 9. 1/[2 + 1/(x+3)] = 1/[2(x+3) +1]/(x+3) = (x+3)/(2x+7) 10. 1/[1 + 1/(a-1)] Which problem in Simmons book is this equivalent to? 5C on page 39, Ans. 1 – 1/a 11. (a – b)/[1/a – 1/b] = (a-b)/[(b-a)/(ab)] = -1ab 12. (x – 1)(x3 + x2 + x + 1) = x4 - 1


Lesson 3 Exercises Evaluate: 1. (23)2 = ?

2. 5453=5n n = ?

3. 5363= n3 n = ?

4. 454-5 = ?

5. a3an = a10 n = ?

6. n3nx = 170 x = ?

7. (ab)c = an n = ? Simplify by removing negative and zero exponents. (x means times or multiply) 8. 24x2163x2-3x7x216-3 = 9. [(a2 + b-2)/(a3 – b2)]0 = Simplify: 10. (5a2b-3)x(2a-3b2) = 11. (x – y)/(x2 – y2) = 12. (x-2 + y-2)(x2 + y2)-1 = 13. (x – 1)(x4 + x3 + x2 + x + 1) =


Lesson 3 Exercise Answers Evaluate: 1. (23)2 = 23x2 = 26 = 64

2. 5453=5n n = 7

3. 5363= n3 n = 30

4. 454-5 = 1

5. a3an = a10 n = 7 Note: 3 + n = 10

6. n3nx = 170 x = -3 Note: 3 + x = 0 .

7. (ab)c = an n = bc Simplify by removing negative and zero exponents. (x means times or multiply) 8. 24x2163x2-3x7x216-3 = 3x7 = 21 9. [(a2 + b-2)/(a3 – b2)]0 = 1 Simplify: 10. (5a2b-3)x(2a-3b2) = 10a-1b-1= 10/(ab) 11. (x – y)/(x2 – y2) = 1/(x+y)

12. (x-2 + y-2)(x2 + y2)-1 = [(y2 + x2)/(xy)2]/(x2 + y2) = 1/(xy)2 13. (x – 1)(x4 + x3 + x2 + x + 1) = x5 - 1


Lesson 4 Exercises Simplify or Calculate (Check with calculator):

1. 3√81 2. √81 3. 4√81 4. 3√(125/27) 5. 4√16 6. 4√(81/625) 7. 6√64 8. 5√100,000 9. (125/27)1/3 = 10. 3√(a3b6) 11. [2 – (√5/2)2]1/2 12. 3√x2 13. √x3

Simplify by rationalizing the denominator:

14. 25/√10 15. (√7 + 3)/( √7 – 3) 16. 3/(√3 + √5 )


Lesson 4 Exercise Answers Simplify or Calculate (Check with calculator) NOTE: You will need to understand the radical version of the answers for the SAT and ACT tests. The calculator number answer is to check your work.

1. 3√81 = 3√(3x27) = 3√3x3 = 4.33

2. √81 = 9

3. 4√81 = 3

4. 3√(125/27) = 5/3

5. 4√16 = 2

6. 4√(81/625) = 3/5

7. 6√64 = 2

8. 5√100,000 = 10

9. (125/27)1/3 = 5/3 see #4

10. 3√(a3b6) = ab2

11. [2 – (√5/2)2]1/2 = [2-(5/4)] ½ = √3/2

12. 3√x2 = x2/3

13. √x3 = x3/2 Simplify by rationalizing the denominator

14. 25/√10 = (25/10)√10=5√10/2=7.906

15. (√7 + 3)/( √7 – 3) = (√7 + 3)2/(-2) = = -8 -3√7 = -15.937 Note: (√7 + 3)( √7 – 3) = 7 – 9 = -2

16. 3/(√3 + √5 ) = 3(√3 - √5 )/(-2)= -(3/2)(√3 - √5) = (3/2)(√5 -√3) = .756

Note: (√3 + √5 )(√3 - √5 ) = 3 - 5 = -2


Lesson 5 Exercises Add or Subtract: 1. (2X6-3X5+6X3+X-7)+(5X5-3X4+6X3+8X-10) 2. (X6-3X5+6X4+8X-17)-(5X5-3X4+6X3+8X-10) 3. (-3X5+6X4+8X-17)-(5X5-3X4+6X3+8X2-10)

Multiply: 4. (3X3+2X2-X+1)(2X2-X+3) 5. (-X3+3X2-X+1)(X2-4X+2) 6. (X2-4X+2)(-X3+3X2-X+1) 7. (X2+1)(X2-1) 8. (X3+1)(X3-1) 9. (Xn+1)(Xn-1) 10. (X2+1)(X2+1) 11. (Xn+1)(Xn+1) 12. (x – 1)(x4 + x3 + x2 + x + 1)


Lesson 5 Exercise Answers Add or Subtract: 1. (2X6-3X5+6X3+X-7)+(5X5-3X4+6X3+8X-10) 2X6+2X5-3X4+12X3+9X-17 2. (X6-3X5+6X4+8X-17)-(5X5-3X4+6X3+8X-10) X6-8X5+9X4-6X3-7 3. (-3X5+6X4+8X-17)-(5X5-3X4+6X3+8X2-10) -8X5+9X4-6X3-8X2+8X -7 Multiply: 4. (3X3+2X2-X+1)(2X2-X+3) = 6X5+X4+5X3+9X2-4X+3 5. (-X3+3X2-X+1)(X2-4X+2) = -X5+7X4-15X3+11X2-6X+2 6. (X2-4X+2)(-X3+3X2-X+1) = -X5+7X4-15X3+11X2-6X+2 7. (X2+1)(X2-1) = X4-1 8. (X3+1)(X3-1) = X6-1 9. (Xn+1)(Xn-1) = X2n-1 10. (X2+1)(X2+1) = X4+2X2+1 11. (Xn+1)(Xn+1) = X2n+2Xn+1 12. (x – 1)(x4 + x3 + x2 + x + 1) = X5-1


Lesson 6 Exercises Factor using integers: 1. X2 – 4 2. X2 + 4 3. X2-X-2 4. 3X2-3X-6 5. X2-2X-8 6. X2+2X-8 7. a2+2a-8 8. X2-6X+9 9. X2+6X+9 10. X2-6X-9 11. X4+6X2+9 12. X3–1 13. X5–1

Hint: xn–1 = (x–1)(xn-1 + xn-2 + . . . + x+1)

14. √3X2-2√3X-8√3 15. X4–1


Lesson 6 Exercise Answers Factor using integers: 1. X2 – 4 = (X-2)( X+2) 2. X2 + 4 Can’t factor with integers

3. X2-X-2 = (X-2)( X+1)

4. 3X2-3X-6 = 3(X-2)( X+1)

5. X2-2X-8 = (X-4)( X+2)

6. X2+2X-8 = (X+4)( X-2) 7. a2+2a-8 = (a-2)(a+4)

8. X2-6X+9 = (x-3)2

9. X2+6X+9 = (x+3)2 10. X2-6X-9 = Can’t factor with integers

11. X4+6X2+9 = (x2+3)2

12. X3–1 = (X – 1)( X 2 + X + 1)

13. X5–1 = (X – 1)( X 4 + X 3 + X 2 + X + 1)

14. √3X2-2√3X-8√3 = √3(X– 4)( X+2)

15. X4–1 = (X2–1)(X2+1) =

(X-1)(X+1)( X2+1) = (X-1)( X 3+X 2+X+1) Note: An important identity you may find useful on tests is: Hint: xn–1 = (x–1)(xn-1 + xn-2 + . . . + x+1) In Tier 4 we will use this to explain the n roots of unity. The solutions to the equation: Xn – 1 = 0 are n complex numbers.


Lesson 7 Exercises Write out the Rule of Algebra. What is NOT permissible? Solve for the obvious unknown or x. You might have to review Algebra from the Practical Math Foundation. It’s pretty amazing how much algebra you already have learned. Just apply the Rules of Algebra you already know, or should remember. 1. 4x + 5 = 0

2. 3x – 4 = 7 – 2x

3. 3 + 4y = 10 + y

4. 3x–6x+8 = 7x–2x+17

5. √x = 4

6. √x = -4

7. X2 = 9

8. X2 = -9

9. 3/x = 5

10. 3/√x = 12

11. 16/X2 = 1/4

12. a/X2 = 1/a

13. (X + 1)2 = 9

14. (2X + 4)2 = 9

15. √(2x+1) = 4

16. (2x+1)1/2 = 4


Lesson 7 Exercise Answers Write out the Rule of Algebra.

If you have an equation LS = RS you may create a new equivalent equation by performing the same mathematical operation on BOTH sides of the original equation.

What is NOT permissible? Division by 0 Solve for the obvious unknown or x. You might have to review Algebra from the Practical Math Foundation. It’s pretty amazing how much algebra you already have learned. Just apply the Rules of Algebra you already know, or should remember. 1. 4x + 5 = 0 x = -(5/4)

2. 3x – 4 = 7 – 2x x = 11/5

3. 3 + 4y = 10 + y y = 7/3

4. 3x–6x+8 = 7x–2x+17 x = -9/8

5. √x = 4 x = 16

6. √x = -4 x = 16

7. X2 = 9 X = 3, or X = -3

8. X2 = -9 No Real Solution

9. 3/x = 5 x = 3/5

10. 3/√x = 12 x = 1/16

11. 16/X2 = ¼ X = 8 or X = -8

12. a/X2 = 1/a X = a or X = -a

13. (X + 1)2 = 9 X = 2 or X = -4

14. (2X + 4)2 = 9 X = -1/2 or X = -7/2

15. √(2x+1) = 4 x = 15/2

16. (2x+1)1/2 = 4 x = 15/2


Lesson 8 Exercises

1. What are the two roots or solutions of: Ax2 + Bx + C = 0 2. What are the two roots or solutions of: Px2 + Qx + S = 0 3. What are the two roots or solutions of: Cx2 + Ax + B = 0 Solve for the solutions or roots of: 4. x2 + x - 2 = 0

5. x2 + x - 3 = 0

6. x2 + 2x - 3 = 0

7. x2 - 2x - 2 = 0

8. 2x2 + 2x - 3 = 0

9. 2y2 + 2y - 3 = 0

10. 3x2 + 2x - 4 = 0

11. 3x2 - 2x + 4 = 0

12. x2 + x + 1 = 0

13. x3 – 1 = 0

14. x5 – 1 = 0

15. 2x4 + 2x2 - 3 = 0 An Honors Student problem!


Lesson 8 Exercise Answers

1. What are the two roots or solutions of: Ax2 + Bx + C = 0 [-B+(B2–4AC)1/2]/2A ; [-B-(B2–4AC)1/2]/2A

2. What are the two roots or solutions of: Px2 + Qx + S = 0

[-Q+(Q2–4PS)1/2]/2P ; [-Q-(Q2–4PS)1/2]/2P 3. What are the two roots or solutions of: Cx2 + Ax + B = 0

[-A+(A2–4CB)1/2]/2C ; [-A-(A2–4CB)1/2]/2C This is why formulas can be confusing. You must be sure what each of the parameters or constants really stand for. #1 is how the quadratic formula is usually displayed. But, these other two would be just as good, just different constants for the coefficients. Solve for the solutions or roots of: 4. x2 + x - 2 = 0 x = -2, x = 1 Could solve by factoring 5. x2 + x - 3 = 0 x = 1/2 + √13/2 or x = 1/2 - √13/2 6. x2 + 2x - 3 = 0 x = -3 or x = 1 Could solve by factoring 7. x2 - 2x - 2 = 0 x = 1 + √3 or 1 - √3 8. 2x2 + 2x - 3 = 0 x = -1/2 + √28/4 or x = -1/2 - √28/4

You might note: √28 = √4x√7 = 2√7 9. 2y2 + 2y - 3 = 0 y = -1/2 + √28/4 or y = -1/2 - √28/4 10. 3x2 + 2x - 4 = 0 x = -1/3 + √52/6 or x = -1/3 + √52/6

You might note: √52 = √4x√13 = 2√13 11. 3x2 - 2x + 4 = 0 No Real Solution


Lesson 8 Exercise Answers continued 12. x2 + x + 1 = 0 No Real Solution. i = √-1

x = -1/2 + √3/2i, or x = -1/2 + √3/2i We will learn all about these “complex numbers” in Tier 4. They are very important in STEM subjects. 13. x3 – 1 = 0 x = 1, plus two solutions in #12. All of these solutions are called the 3rd Roots of Unity since they all satisfy x3 = 1 14. x5 – 1 = 0 x = 1, and… …the other four 5th roots of unity, which cannot be found with the quadratic formula. This problem is fairly easy to solve with trigonometry when it is extended to complex numbers in Tier 4. And now the problem for “honors” students. You might run into such a question on the SAT or ACT, but I doubt it. 15. 2x4 + 2x2 - 3 = 0

Hint: let y = x2 and solve for y first, then for x. 2y2+2y-3=0 See Problem # 9. Note the second root is negative and thus has no real square root. So we have two solutions now:

x = [-1/2+√28/4]1/2 and x = -[-1/2+√28/4]1/2 If there had been two positive roots, then there would have been four real number solutions. Example: (x-2)(x-3)=x2 – 5x + 6 = 0 has two roots, x =2 and x = 3. So x4 – 5 x2 + 6 = 0 will have four real roots:

x = √2, x = -√2, x = √3, x = -√3


Lesson 9 Exercises Evaluate: 1. │-3│=

2. │3│=

3. │7-3│=

4. │-7+3│= Determine all x for which the inequality will be true: 5. 3x + 4 < 0

6. 3x + 4 < 10

7. 3x + 4 < -10

8. 3x + 4 < 2x + 7

9. 3x + 4 < -2x + 7

10. 3x + 4 > -2x + 7

11. 6 – 7x > 4x – 8

12. 6 – 7x < 4x – 8

13. │x│< 1

14. │x│< -1

15. │x│< 10

16. │x + 2│< 1

17. (x-2)(x+3)<0

18. (x-2)(x+3)>0

19. X2+4x-5<0

20. X2+4x-5>0


Lesson 9 Exercise Answers Evaluate. 1. │-3│= 3

2. │3│= 3

3. │7-3│= 4

4. │-7+3│= 4 Determine all x for which the inequality will be true. 5. 3x + 4 < 0 x < -4/3

6. 3x + 4 < 10 x < 2

7. 3x + 4 < -10 x < -14/3

8. 3x + 4 < 2x + 7 x < 3

9. 3x + 4 < -2x + 7 x < 3/5

10. 3x + 4 > -2x + 7 x > 3/5

11. 6 – 7x > 4x – 8 x < 14/11

12. 6 – 7x < 4x – 8 x > 14/11

13. │x│< 1 -1 < x < 1

14. │x│< -1 No Solutions

15. │x│< 10 -10 < x < 10

16. │x + 2│< 1 -3 < x < -1

17. (x-2)(x+3)<0 -3 < x < 2

18. (x-2)(x+3)>0 x < -3 and x > 2

19. X2+4x-5<0 -5 < x < 1

20. X2+4x-5>0 x < -5 and x > 1

Tier 3 Part 1 Exercises 2014 - Triad...

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