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CST Prep Part II MAIN MENU Standard 10 Standard 11 Standard 12 Standard 13 Standard 14 Standard 15 Standard 16 Standard 17 Standard 18 Standard 19 Standard 20 Standard 21 Standard 22 Standard 23 Standard 25.1 Designed by Ms.Carranza and Mrs. Murray Solved by: 8 th Grade Gate Students 2011

CST Prep Part II MAIN MENU

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CST Prep Part II MAIN MENU. Designed by Ms.Carranza and Mrs. Murray Solved by: 8 th Grade Gate Students 2011. Standard 10.0 Students add, subtract, multiply, and divide monomials and polynomials. Students solve multi-step problems, including word problems, by using these techniques. - PowerPoint PPT Presentation

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Page 1: CST Prep Part II MAIN MENU

CST Prep Part IIMAIN MENU

Standard 10

Standard 11

Standard 12

Standard 13

Standard 14

Standard 15

Standard 16

Standard 17

Standard 18

Standard 19

Standard 20

Standard 21

Standard 22

Standard 23

Standard 25.1

Designed by Ms.Carranza and Mrs. Murray Solved by: 8th Grade Gate Students 2011

Page 2: CST Prep Part II MAIN MENU

Standard 10.0 Students add, subtract, multiply, and divide monomials and polynomials.

Students solve multi-step problems, including word problems, by using these techniques.

Problem 47 Problem 49

Problem 51

Problem 48

Problem 50Main Menu

Problem 52

Page 3: CST Prep Part II MAIN MENU

51) A volleyball court is shaped like a rectangle. It has a width of x meters and length of 2x meters. Which

expression gives the area of the court in the square meters?

A 3xB 2x²C 3x²D 2x³

Vocabulary Rules & Strategies

Solution & Answer Standard 10

Page 4: CST Prep Part II MAIN MENU

Vocabulary

• Expression : a mathematical phrase that contains operations, numbers, and/or variables.

• Area: The number of non overlapping unit squares of a given size that will exactly cover the interior of a plane figure.

• Square Meters: a unit of area measurement equal to a square measuring one meter on each side.

Back to Problem

Page 5: CST Prep Part II MAIN MENU

Rules & Strategies

• Area= base (length) * height (width)• Product of Powers: combine base, add

powers• Add exponents ( x = x )

Back to Problem

1

x

2x

Page 6: CST Prep Part II MAIN MENU

Solution

• 1x

2x

Answer: BBack to Problem

Standard 10

2x

2

1 1

Page 8: CST Prep Part II MAIN MENU

Solution and Answers (x+2)+(x-2x)(2x+1)?

Step 1: Area Method

Step 2: Combine like terms and put in Descending order

Answer: 2x² - 2x

2x + 12x + 1 xx -2 2x²-3x-2-2 2x²-3x-2

1X + 2 + 2x² - 3x - 21X + 2 + 2x² - 3x - 2

2x²2x² +x+x

-4x-4x -2-2

-2x-2x2x²2x²0+0+

Back to Problem

Standard 10

Page 9: CST Prep Part II MAIN MENU

Vocabulary

Area Method: Area Method: ax + bax + b

(ax + b)(ax + b) (ax + b)(ax + b) axax b b aax aax + + (abx + abx) (abx + abx) + + bbbb

Descending Order: Descending Order: ordering terms from greatest to least((ex. 1x+ 2x² + 3x+1x³ + 9 = 1x³ + 2x² + 3x + 1x + 9)ex. 1x+ 2x² + 3x+1x³ + 9 = 1x³ + 2x² + 3x + 1x + 9)

abxabx

aaxaax abxabx

bbbb

Back to Problem

Page 11: CST Prep Part II MAIN MENU

49)The sum of the two binomials is 5x2-6x, If one of the binomials is 3x2-2x, what is the

other binomial

A 2x2-4xB 2x2-8xC 8x2+4xD 8x2-8x

Vocabulary Rules & Strategies

Standard 10Solution and

Answer

Page 12: CST Prep Part II MAIN MENU

Vocabulary

• sum– answer t an addition problem • Binomial- A polynomial with two terms

Back to Problem

Page 13: CST Prep Part II MAIN MENU

Rules and strategies

• Line up the binomials• Combine both binomials by subtracting

Back to Problem

Page 14: CST Prep Part II MAIN MENU

Solution

(3x2-2x)+2x2-4x5x2-6x

Answer: A Standard 10

1st binomial

2nd binomial

Sum of binomials

Back to Problem

Page 15: CST Prep Part II MAIN MENU

47) 5x3

10x7

A) 2x4

B)1 2x4

C) 1 5x4

D) X4

5

Vocabulary Rules & Strategies

Standard 10.0Solution and

AnswerSolution and

Answer

Page 16: CST Prep Part II MAIN MENU

Vocabulary

• Quotient of Powers: simplify coefficientscombine base

subtract exponents

Back to Problem

Page 17: CST Prep Part II MAIN MENU

Rules & Strategies

• In order to skip the negative exponent rule, circle the biggest exponent to tell if the variable stays in the denominator or numerator.

• Subtract powers • Divide if there is any whole numbers in the

numerator and denominator.

Back to Problem

Page 18: CST Prep Part II MAIN MENU

Solution

5x3 1

10x7 2x4

Answer: B Back to Problem Standard 10.0

Page 19: CST Prep Part II MAIN MENU

48.) (4x²-2x+8) – (x²+3x-2)

Vocabulary Rules and Strategies

Solution and Answer Standard 10

Page 20: CST Prep Part II MAIN MENU

Vocabulary

• Parenthesis ( ); indicates separate grouping of symbols

• Exponent ²; a symbol or number placed above and after another symbol or number to denote the power to which the latter is to be raised

• Variable ‘x’; a quantity or function that may assume any given value or set of values

Back to Problem

Page 21: CST Prep Part II MAIN MENU

Rules and Strategies

• Subtracting Polynomials / Lesson 7-7• Distribute (-) before grouping

Back to Problem

Page 22: CST Prep Part II MAIN MENU

Solution and Answer

Answer:D Back to Problem Standard 10

(4x²-2x+8) – 1 (x²+3x-2)(4x²-2x+8)-x²-3x+2

(4x²-1x²) + (-2x-3x) + (8+2)

3x²-5x+10

Page 23: CST Prep Part II MAIN MENU

Standard 11 Students solve multistep problems involving linear equations and

inequalities in one variable.

Problem 54

Problem 56

Problem 53

Problem 55

Main Menu

Page 24: CST Prep Part II MAIN MENU

53)Which is the factored form of 3a²-24ab+48b²?

a. (3a-8b)(a-6b)b. (3a-16b)(a-3b)c. 3(a-4b)(a-4b)d. 3(a-8b)(a-8b)

Vocabulary Rules & Strategies

Solution & Answer Standard 11

Page 25: CST Prep Part II MAIN MENU

Vocabulary

• Factored form= Form of equation in which each term is simplified from factoring methods and GCF.

Back to Problem

Page 26: CST Prep Part II MAIN MENU

Rules & Strategies

• Ask yourself if you can factor out a GCF• Use the diamond method• Keep asking yourself whether you can factor more.If there is a GCF, don’t forget to Include it in your final answer.

Back to Problem

Page 27: CST Prep Part II MAIN MENU

Solution & Answer

Factor out a gcf3a²-24ab + 48b²

3 3 3 3(a²-8ab+16b²) +16

keep factoring!! a a -4b -4b

3 (a-4b)2

Solution: C -8 Back to Problem

Standard 11

Page 28: CST Prep Part II MAIN MENU

54) Which is the factor of x² – 11x + 24

A x + 3 B x - 3

C x + 4 D x - 4

Vocabulary Rules & Strategies

Solution & Answer Standard 11

Page 29: CST Prep Part II MAIN MENU

Vocabulary

• Factor = A number or expression that is multiplied by another number or expression to get a product.

• Term = a part of an expression that is added or subtracted

• Binomial = 2 terms• Trinomial = 3 terms

Back to Problem

Page 30: CST Prep Part II MAIN MENU

Rules & Strategies• Since there are three terms in this trinomial you use the “Diamond Method”• Diamond Method Formula : ax² + bx + c• Make an “X” on your paper.• On the top intersect write a multiplication sign (x) to show that the two

denominators multiply to equal the number you're going to get above the multiplication sign.

• On the bottom intersect write a plus sign (+) to show that the two denominators added together equal the number you’re going to get below the addition sign.

• Plug the value for “c” above the multiplication sign ( +24 )• Plug in the value for “b” below the addition sign ( -11 )• Now , think of two numbers that multiplied equal +24 and added equal -11 . • ( You should come up with -3 and -8 )• Lastly , you rewrite your fractions as binomials : 1x = ( x – 3 ) -3

Back to Problem

Page 31: CST Prep Part II MAIN MENU

Solution

x² – 11x + 24 + 24 1x x 1x -3 + -8 - 11 ( x – 3 ) ( x – 8 ) Final answer

Answer: BBack to Problem Standard 11

Page 32: CST Prep Part II MAIN MENU

55) Which of the following shows 9t + 12t + 4 factored completely?

A (3t + 2) B (3t + 4) (3t +1) C (9t + 4) (t + 1)

D 9t + 12t + 4

Vocabulary Rules & Strategies

Solution & Answer

2

2

2

Standard 11

Page 33: CST Prep Part II MAIN MENU

Vocabulary

• Factored :one of two or more numbers, algebraic expressions, or the like, that when multiplied together produce a given product; a divisor.

Back to Problem

Page 34: CST Prep Part II MAIN MENU

Rules & Strategies

• Always ask yourself which factoring method should I use?

• You should always see if you can factor out a GCF

• Use“diamond method” when there is no GCF and the coefficient is greater than one.

Back to Problem

Page 35: CST Prep Part II MAIN MENU

Solution

9t + 12t + 4

+36 3 9t 9t 3 3t 3 +6 +6 3 2

+12

(3t+2) Back to

Problem

2

3t 2

•Make sure to simplify the fractions

2Answer: A

Standard 11

Page 36: CST Prep Part II MAIN MENU

56) What is the complete factorization of 32-8z²

A. -8(2+z)(2-z)B. 8(2+z)(2-z)

C. -8(2+z)²D. 8(2-z)²

Vocabulary Rules & Strategies

Standard 11Solution &

Answer

Page 37: CST Prep Part II MAIN MENU

Vocabulary

• Factorization: to simplify

Back to Problem

Page 38: CST Prep Part II MAIN MENU

Rules & Strategies

• Find GCF• Divide each term by GCF• Rewrite equation w/ GCF outside of parenthesis• Divide inside terms by a negative • Rewrite the equation w/ negative on the outside• Difference of 2 Squares• Rewrite new equation & THAT’S YOUR ANSWER !

Back to Problem

Page 39: CST Prep Part II MAIN MENU

Solution

32-8z²-8 -8

factor out a -8 to get a positive z²-8(z²-4) keep factoring diff of 2 squares

-8(z+2)(z-2)Or

-8(2+z)(2-z)

Answer: A. -8(2+z)(2-z) Back to Problem

Standard 11

Page 40: CST Prep Part II MAIN MENU

Standard 12

• Students simplify fractions with polynomials in the numerator and denominator by factoring both and reducing them to the lowest terms.

Problem 78 Problem 79 Problem 80

Problem 81Main Menu

Problem 77

Page 41: CST Prep Part II MAIN MENU

78.) 6x2 + 21x + 9 4x2 - 1

Vocabulary Rules and Strategies

Solution and Answer Standard 12

Page 42: CST Prep Part II MAIN MENU

Vocabulary

• GCF ; Greatest Common Factor • Super Diamond ; x2 + bx + x• Difference of Two Squares ; terms need to be in

perfect squares [Example: (a+b)(a-b)]

Back to Problem

Page 43: CST Prep Part II MAIN MENU

Rules and Strategies

• Top ; find GCF, super diamond• Bottom ; difference of two squares, terms

need to be perfect squares

Back to Problem

Page 44: CST Prep Part II MAIN MENU

Solution and Answer

Back to Problem Standard 12

6x2 + 21x + 9

Top/Numerator ; 6x2: 2 ∙ 3 ∙ x ∙ x21x: 3 ∙ 7 ∙ x 9: 3 ∙ 3

GCF: 3

6x 2 +21 + 9 3 3 33 ( 2x 2 + 7x + 3)

Super Diamond!!

1) GCF2) Difference of Two Squares3)Cross Cancel

∙+

+6

+7

+2x+6

+2x+1

x

+3

3(x+3)(2x+1)

Bottom/Denominator ;4x2 - 1

2x 2x 1 1(2x+1)(2x-1)

Top/Numerator ;

6x2 + 21x + 9 4x2 – 1

Cross Cancel3(x+3)(2x+1)(2x+1)(2x-1)

=Answer ; B 3(x+3) 2x-1

=

Page 45: CST Prep Part II MAIN MENU

79)What is x2-4x+4 reduced to lowest terms?

x2-3x+2A) x-2 X-1B) x-2 X+1C) x+2 X-1D) x+2 X+1

Vocabulary Rules & Strategies

Solution & Answer Standard 12

Page 46: CST Prep Part II MAIN MENU

Vocabulary

• Reduce = lower in degree• Lowest term = the form of a fraction after

dividing the numerator and denominator by their greatest common divisor.

Back to Problem

Page 47: CST Prep Part II MAIN MENU

Rules & Strategies

• First you should ask yourself which strategy is best to change the equation into a dividable state using either; Greatest common facture (GCF), difference of 2 squares, or one of the diamond methods.

• In this case the best one would be the diamond method on both the denominator and numerator.

• Once you find the factors of both of the trinomials reduce by dividing the like terms.

Back to Problem

Page 48: CST Prep Part II MAIN MENU

Solution ax2+bx+c=0

Step 1: x2-4x+4 x2-3x+2

Step 2: x2-4x+4 (x-2) (x-2) x2-3x+2 (x-2) (x-1) Diamond Method

4 2 x * x x * x -2 -2 -2 -1 + +

-4 Divide out! -3

(x-2) (x-2) = (x-2) Answer: A (x-2) (x-1) (x-1) Back to

Problem Standard 12

Page 49: CST Prep Part II MAIN MENU

80. What is 12a3 – 20a2 reduced to lowest terms? 16a2 + 8a

Vocabulary Rules and Strategies

Solution and Answer Standard 12

Page 50: CST Prep Part II MAIN MENU

Vocabulary

• Reduced ; simplified or lowest terms

Back to Problem

Page 51: CST Prep Part II MAIN MENU

Rules and Strategies

• Find GCF for both numerator and denominator

• Simplify to lowest terms

Back to Problem

Page 52: CST Prep Part II MAIN MENU

Solution and Answer 12a3 - 20a2 GCF:4a2 16a2 + 8a GCF: 8a2

4 a2 (3a – 5) *Simplify 8 a (2a + 1)

a (3a – 5)2 (2a +1)

Answer:D Back to Problem Standard 12

Numerator :12a3 - 20a2

4a2 4a2

4a2 (3a – 5)

Denominator:16a2 + 8a 8a 8a8a (2a + 1)

Page 53: CST Prep Part II MAIN MENU

81) What is the simplest for of the fraction: _x2-1_ x2+x-2

A)_-1__ x-2B)x-1 x-2C)x-1 x+2D)x+1 x+2

Vocabulary Rules & Strategies

Solution & Answer Standard 12

Page 54: CST Prep Part II MAIN MENU

Vocabulary

• Simplest form of a rational expression- A rational expression is in simplest form if the numerator and denominator have no common factors. Ex:

x2-1 (x-1)(x+1)x2+x-2 (x+1)(x+2) (x+1) (x+2) Simplest Form Back to

Problem

Page 55: CST Prep Part II MAIN MENU

Rules & Strategies

• First ask yourself: “1)How will I factor?2)Simplify/ Divide out

*make sure to separate work to make your work easier to read.

Back to Problem

1.GCF2.Diamond

3.SuperDiamond4.Difference of 2 Squares

Page 56: CST Prep Part II MAIN MENU

Solution

Answer:DBack to Problem Standard 12

_x2-1_ x2+x-2

WORK:Step 1) difference of 2 squares/diamond

Difference of 2 squares:

x2- 1x x 1 1 (x-1)(x+1) Diamond:

x x-1 +2

+1

-2

(x-1)(x+2)

(x-1)(x+1)(x-1)(x+2)

Page 57: CST Prep Part II MAIN MENU

Standard 13 Students add, subtract, multiply, and divide rational expressions

and functions. Students solve both computationally and conceptually challenging problems by using these techniques.

Problem 82 Problem 84Problem 83

Problem 85 Main Menu

Page 58: CST Prep Part II MAIN MENU

82) 7z²+7z • z²-4 4z+8 z³+2z²+z

a. 7(z-2) b.7(z+2) 4(z+1) 4(z-1)c. 7z(z+1) d. 7z(z-1) 4(z+2) 4(z+2)

Vocabulary Rules & Strategies

Solution & Answer

Standard 13

Page 59: CST Prep Part II MAIN MENU

Vocabulary

Factoring= The process of writing a number or algebraic expression as a product.

GCF= For two or more numbers, the largest whole number that divides evenly into each number.

Back to Problem

Page 60: CST Prep Part II MAIN MENU

Rules & Strategies

Check whether to use GCF, before choosing a factoring method.

Divide out common binomials.

Back to Problem

Page 61: CST Prep Part II MAIN MENU

Solution & AnswerGCF:7z -> 7z²+7z • z²-4 <- Difference of 2 SquaresGCF: 4 -> 4z+8 z³+2z²+z <- GCF: z; Diamond

7z(z+1) • (z+2)(z-2) 4(z+2) z (z+1) (z+1)

7z(z+1) • (z-2) 4 z (z+1) (z+1)

Answer: A = 7(z-2) 4(z+1) Back to

Problem

Standard 13

Page 62: CST Prep Part II MAIN MENU

84) x+8x+16 2x+8 x+3 x-9

A 2(x+4) (x-3)(x+3)B 2(x+3)(x-3) x+4

Vocabulary Rules & Strategies

Solution & Answer part 1 Standard 13

2

2

C (x+4)(x-3) 2D (x+4)(x-3) 2(x+3)

2

22

Page 63: CST Prep Part II MAIN MENU

Vocabulary

Rational Expression: a quotient of polynomials

Back to Problem

Page 64: CST Prep Part II MAIN MENU

Rules & Strategies

• Take the reciprocal of the fraction on the right side of the sign and then multiply.

• Factor each term if needed.• Divide out common factors, if needed.

Back to Problem

Page 65: CST Prep Part II MAIN MENU

Solutionx+8x+16 2x+8 x+8x+16 x-9 x+3 x-9 x+3 2x+8x+8x+16 2x+8 x-9 2x: 2 x x x 3 3 (x+4)(x+4) (x+3)(x-3)

8: 2 2 2 (x+3) (x-3) x+3 2(x+4)

Diivide out!

(x+4) (x-3)

2 answer: C

Back to

Problem Standard 13

2

2

2 2

Greatest Common Factor: 2

Difference of 2

+16

+8

+

.+4 +41x 1x

(x+4)

GCF: 2

2x+822

2(x+4)

2

Change to multiplication and take reciprocal of second rational

expression

Change to multiplication and take reciprocal of second rational

expression

LOOK BEFORE YOU BEGIN!

2

W

R

O

K

Page 66: CST Prep Part II MAIN MENU

85) Which fraction is equivalent to 3x5

x x4 + 2

A. X² 5B. 9x² 20

c. 4 5D. 9 5

Vocabulary Rules &

Strategies

Solution & Answer Standard 13

Page 67: CST Prep Part II MAIN MENU

Vocabulary

• Fraction: a ratio of two expressions or numbers other than zero

• Equivalent: equal

Back to Problem

Page 68: CST Prep Part II MAIN MENU

Rules & Strategies

• Find the least common denominator (LCD)• Change division into multiplication• With the second fraction, take the reciprocal• Cross cancel• Multiply straight across

Back to Problem

Page 69: CST Prep Part II MAIN MENU

Solution & Answer3x5

x x4 + 2

x x 4 = 4

x 2x+ 2 = 4

3x take reciprocal

4Answer: C. 3x 4 4 5 3x 5 Back to

Problem

Standard 13

find Common denominator = 4

find Common denominator = 4

Page 70: CST Prep Part II MAIN MENU

83) Which fraction equals the product?

A 2x -3 3x+2

B 3x+2 4x-3

C x²-25 6x²-5x-6

D 2x²+7x-15 3X²-13X-10

Vocabulary Rules & Strategies

Solution & Answer Standard 13

Page 71: CST Prep Part II MAIN MENU

Vocabulary

• Fraction = a ratio of algebraic quantities similarly expressed

• Product = the answer of a multiplication problem • Area Method = multiply outside add or subtract

inside terms (like terms)• Numerator = the top numbers of a fraction• Denominator = the bottom numbers of a fractionBinomials: two terms Back to

Problem

Page 72: CST Prep Part II MAIN MENU

Rules & Strategies• Since there are two different fractions in parenthesis and both the numerators and

denominators are binomials you use “Area Method”• Area method is when you draw out a rectangle and divide it into 4 quadrants (4 squares)• Then, you put one binomial on the left side written out like this:

• Next, you put the other binomial adjacent (next to) the fraction on top of the rectangle • Then you multiply each term to each other and Place each number in a square (label them 1, 2, 3, and 4)• Write out numbers 1 and 4 into an expression as the first andlast terms : 2x² + 15• Now, since numbers 2 and 3 are like terms combine them and putthem in between 2x² and +15 : 2x² +7x +15• Repeat this process to the other binomial and place the expression underneath 2x² +7x +15 .

Back to Problem

Page 73: CST Prep Part II MAIN MENU

Solution

Answer: D Back to Problem Standard 13

3x²-13x-10

+5

+2x -3

2x²

10x

-3x

-15

²

-5+x

+3x

+2

3x² -15x

2x -10

2x²+7x-15 Combine2x²+7x-15 on the top and 3x²-13-10on the bottom

+x

Page 74: CST Prep Part II MAIN MENU

Standard 14 Students solve a quadratic equation by factoring or completing

the square.

Problem 57 Problem 59

Problem 61

Problem 58

Problem 62

Problem 60 Main Menu

Page 75: CST Prep Part II MAIN MENU

59) What are the solutions for the quadratic equation x + 6x = 16?

A -2,-8B -2, 8C 2,-8D 2,8

Vocabulary Rules & Strategies

Solution & Answer Standard 14

2

Page 76: CST Prep Part II MAIN MENU

Vocabulary

• Quadratic equation : an equation containing a single variable of degree 2.

Back to Problem

Page 77: CST Prep Part II MAIN MENU

Rules & Strategies

• the “diamond method”• All the numbers have to be on one side of the

equation.

Back to Problem

Page 78: CST Prep Part II MAIN MENU

Solution

ax + bx + c = 0x +6x = 16

-16 -16 x + 6x – 16 =0

(x + 8) (x - 2) = 0 Solve each factor to = 0

x + 8 = 0 x – 2 = 0 - 8 - 8 +2 +2 x = -8 x = 2

Back to Problem Standard 14

-16

+6

x-2

x+8

Answer: C

2

2

2

Page 79: CST Prep Part II MAIN MENU

57) If x is added to x, the sum is 42. Which of the following could be the

value of x?

A -7B -6C 14D 42

Vocabulary Rules & Strategies

Solution & Answer Standard 14

2

Page 80: CST Prep Part II MAIN MENU

Vocabulary

• Value: A number represented by a figure, symbol; the value of x.

• Sum: a series of numbers to be added up

Back to Problem

Page 81: CST Prep Part II MAIN MENU

Rules & Strategies

• Plug in the value of x• Set up the expression (x) + x

Back to Problem

2

Page 82: CST Prep Part II MAIN MENU

Solution

Try x=-7 (x) + x= 42

(-7) +(-7)= 42 49 + (-7)=42

42=42 Answer: A Back to

Problem Standard 14

2

2

Page 83: CST Prep Part II MAIN MENU

62) What are the solutions for the quadratic equation x2-8x=9?

A) 3B) 3,-3C) 1,-9D) -1,9

Vocabulary Rules & Strategies

Solution & Answer Standard 14

Page 84: CST Prep Part II MAIN MENU

Vocabulary

• Solutions - the answer itself• Quadratic - involving the square and no higher

power of the unknown quantity; of the second degree.

• Equation - an expression or a proposition, often algebraic, asserting the equality of two quantities.

Back to Problem

Page 85: CST Prep Part II MAIN MENU

Rules & Strategies

• First get into Quadratic Form ax2+bx+c=0.• Then use the diamond method• Finally separate the two factors to = 0.• Solve for “X”

Back to Problem

Page 86: CST Prep Part II MAIN MENU

Solution ax2+bx+c=0

Step 1: x2-8x=9 Step 3: (x-9) (x+1) = 0 Zero Product -9 -9 X-9=0 x+1=0 Property +9=9 -1=-1 Step 2: 1x2-8x-9=0 x=9 x=-1 Answers Diamond Method

-9 x * x +1 -9 +

-8

Answer: DBack to Problem Standard 14

Page 87: CST Prep Part II MAIN MENU

58) What quantity should be added to both sides of this equation to complete

the square?x²-8x=5

A. 4B. -4C. 16D. -16

Vocabulary Rules & Strategies

Solution Standard 14

Page 88: CST Prep Part II MAIN MENU

Vocabulary

• Quantity: amount

Back to Problem

Page 89: CST Prep Part II MAIN MENU

Rules & Strategies

• Identify “b”• Plug in “b” to the equation: b ²

2• Solve to complete the square.• The solution to your equation is the answer

Back to Problem

Page 90: CST Prep Part II MAIN MENU

Solution

b=-8 -8 ²

2(-4)²+16

Answer: C. 16Back to Problem

Standard 14

Page 91: CST Prep Part II MAIN MENU

60) Leanne correctly solved the equation x² + 4x = 6 by completing the square. Which equation is part of her solution?

A ( x + 2 )² = 8 B ( x + 2)² = 10 C ( x + 4 )² = 10 D ( x + 4)² = 22

Vocabulary Rules & Strategies

Solution & Answer Standard 14

Page 92: CST Prep Part II MAIN MENU

Vocabulary• Completing the square : a method for solving

quadratic equations by using steps such as1. Making sure you have the formula x² + bx + c 2. Find b ² 23. Completing the square with the answer to by adding it to both sides4. Factor the left and simplify the right 5. Use Square root property then you’re done Back to

Problem

Page 93: CST Prep Part II MAIN MENU

Rules & Strategies

• Use the factoring method: completing the square

Back to Problem

Page 94: CST Prep Part II MAIN MENU

Solution

x² + 4x = 6

= b = 4 = 2 ² = 4 add to both sides of the equation

2 2 Perfect Square Trinomial : X² + 4x + 4 = 6 + 4 factor the right, simplify the left factor by diamond methodFinal Answer : ( x + 2 )² = 10 take square root of both sides

Answer: BBack to Problem Standard 14

Completing the Square

Completing the Square

Page 95: CST Prep Part II MAIN MENU

61) Carter is solving this equation by factoring.

10x²-25x+15=0Which expression could be one of his correct

factors?a. x+3b.x-3

c.2x+3d. 2x-3

Vocabulary Rules & Strategies

Solution & Answer

Standard 14

Page 96: CST Prep Part II MAIN MENU

Vocabulary

Expression does not have an = Factors can also be the binomials you get when you diamond

Back to Problem

Page 97: CST Prep Part II MAIN MENU

Rules & Strategies

• Ask yourself which factoring method do I use?• GCF? Diamond? Super Diamond? Diff of 2

Squares• Make sure to find two terms which can be

multiplied to the top, but added to the bottom of the diamond.

• Keep asking yourself whether to factor more or not.

Back to Problem

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Solution & AnswerGCF: 5

10x²-25x+15=0 5 5 5 5

5 ( 2x² -5x +3) =0 +6 5( x-1) (2x-3)=0 x = 2x 2x -1 -2 -5 -3 These are all factors, but

they are just asking for one of them.Answer: D Back to

Problem Standard 14

Page 99: CST Prep Part II MAIN MENU

Standard 15 Students apply algebraic techniques to solve rate problems, work

problems, and percent mixture problems.

Problem 85

Problem 91

Problem 87

Problem 89

Problem 86

Problem 90

Problem 88 Main Menu

Page 100: CST Prep Part II MAIN MENU

87) Andy’s average driving speed for a 4-hour trip was 45 miles per hour. During the first 3 hours he drove 40 miles per

hour. What was his average speed for the last hour of his trip?

A 50 miles per hourB 60 miles per hourC 65 miles per hourD 70 miles per hour

Vocabulary Rules & Strategies

Solution& Answer Standard 15

Page 101: CST Prep Part II MAIN MENU

Vocabulary

• Rate = A ratio that compares two quantities measured in different units.

• Average = The sum of the values in a data set divided by the number of data values. Also called the mean.

Back to Problem

Page 102: CST Prep Part II MAIN MENU

Rules & Strategies

• Set up formula distance = rate time• Put the same units together

Back to Problem

.

Page 103: CST Prep Part II MAIN MENU

Solutiond = 45 4d = 180

180 = r 3 isolate rate 3 360 = r

Answer: B Back to Problem Standard 15

.

.

Page 104: CST Prep Part II MAIN MENU

86) A pharmacist mixed some 10%-saline solution with some 15%- saline solution to obtain 100mL of a

12%-saline solution. How much of the 10%-saline solution did the pharmacist use in the mixture?

A) 60mLB) 45mLC) 40mL

D) 25mL

Solution & Answer

Solution & Answer

VocabularyVocabularyRules &

StrategiesRules &

Strategies

Standard 15

Standard 15

Page 105: CST Prep Part II MAIN MENU

VocabularyDistributive PropertyDistributive Property: is when you

distribute what is outside of the parentheses.

Isolate the variable: Isolate the variable: is to simplify the equation using operations to get the

variable alone , on one side of the equal sign.

Back to ProblemBack to Problem

Page 106: CST Prep Part II MAIN MENU

x + y Solution

=100mL

x+y= 100 isolate y -> y=(100-x) Work 0.10x+0.15y=12 -0.15x0.10x+ 0.15(100-x)=12 -0.05x=-3 +0.10x 0.10x+15-0.15x=12 -0.05 -0.05 -0.05x -15 -15

0.10x-0.15x= -3 Answer: A

-0.05x=-3 isolate xBack to

ProblemBack to

Problem

10

15

12%

10% + 15% = 12%

x + y = 100mL

.10x + .15y = .12(100)=12

X=60mLX=60mL

Standard 15

100 mL

Page 107: CST Prep Part II MAIN MENU

Rules & Strategies

• Make a drawing with the values

• Remember that you need to go two spaces to the left when changing a value to a decimal

• Make a system of equation and use substitution

Back to ProblemBack to Problem

Page 108: CST Prep Part II MAIN MENU

88) One pipe can fill a tank in 20 minutes. While another pipe takes

30 minutes to fill the same tank. How long would it take the two pipes together to fill the tank ?

a. 50minb. 25minc. 15mind. 12min

Vocabulary

Solution and Answer

Rules and Strategies

Standard 15

Page 109: CST Prep Part II MAIN MENU

Vocabulary

• None available

Back to Problem

Page 110: CST Prep Part II MAIN MENU

Rules & StrategiesFind the number of minutes it takes to fill the tank

together. There are 60 minutes in 1 hourPut information into fraction (over 1) and find the LCM.

Once LCM is found, distribute the LCM inside whatever is in the parentheses .

Isolate “m” and your answer will be found.

Back to Problem

Page 111: CST Prep Part II MAIN MENU

Solution 1 1 20m + 30m = 1 m=minutes

LCM: 20: 22 5 Choose one with the greatest power 30: 2 5 3LCM: 22 . 5 . 3 = 60 Distribute 60 to original equation 60 1 1 20m + 30m = 1(60) 60 60 20m + 30m= 60 Reduce Fractions 3m + 2m= 60 5m= 60 5 5

Answer: D m=12

Back to Problem Standard 15

Page 112: CST Prep Part II MAIN MENU

89) Two airplanes left the same airport traveling in opposite directions. If one airplane averages 400 miles per hour and

the other airplane averages 250 miles per hour, in how many hours will the distance between the two planes be 1625

miles?

A 2.5B 4C 5

D 10.8

Vocabulary Rules & Strategies

Solution & Answer Standard 15

Page 113: CST Prep Part II MAIN MENU

Vocabulary

• Distance= rate * time isolate time • d= rt r r t= d r

Back to Problem

Page 114: CST Prep Part II MAIN MENU

Rules & Strategies

• Remember to add both planes’ averages • Divide answer to total distance

Back to Problem

Page 115: CST Prep Part II MAIN MENU

Solution 1st plane 2nd plane

avg 400 avg 250 400 + 250 = 650 is the rate time = 1625 (distance)

650(rate) time= 2.5 hours

Back to Problem Standard 15

Page 116: CST Prep Part II MAIN MENU

90) Lisa will make punch that is 25% fruit juice by adding pure fruit juice to a 2-liter mixture that is 10% pure fruit

juice. How many liters of pure fruit juice does she need to add?

A. 0.4 literB. 0.5 literC. 2 litersD. 8 liters

Vocabulary Rules & Strategies

Solution Standard 15

Page 117: CST Prep Part II MAIN MENU

Vocabulary

• Pure=100% or .10• Liters=amount

Back to Problem

Page 118: CST Prep Part II MAIN MENU

Rules & Strategies

• Use substitution• Goal: solve for y• First, isolate “x” then plug it in the 2nd

equation• NOTE: 2 liters of 10% is 2(.10)= 0.2

Back to Problem

Page 119: CST Prep Part II MAIN MENU

Solution and Answerx= 25 % = .25 y= 100% because its it's PURE fruit juice = 1 Solve for y

Results in 2 liters of 10% = (.10) fruit juiceFirst, isolate x then plug it in.x-y = 2 .25x - 1y = 2(.10)

x-y=2-y=2-x (2.4) – y = 2-1 -1 -1 -2.4 -2.4y= -2+x -y = -.4

-1 -1.25x-(-2+x)=.2.25x+2-x=.2 y= .4 -2 -2.25x-x= -1.8

-.75x = -1.8-.75= -.75 x = 2.4

Back to Problem

Standard 15

Page 120: CST Prep Part II MAIN MENU

91) Jenna's car averaged 30 miles per gallon of gasoline on her trip. What is the value of x in gallons of gasoline?

A 32B 41 C 55 D 80

Vocabulary Rules & Strategies

Solution & Answer Standard 15

Miles traveled

600 450 300 960

Gallons of gasoline

20 15 10 x

Page 121: CST Prep Part II MAIN MENU

Vocabulary• Average = The sum of all the values in a data set

divided by the number of data values. Also called the mean.

• Value = The amount of ; or equivalency of .• Rate = A ratio that compares two quantities

measured in different units.• Ratio = A comparison of two numbers of

quantities .• Equation = A mathematical statement that two

expressions equal.Back to Problem

Page 122: CST Prep Part II MAIN MENU

Rules & Strategies

• Rewrite the problem as an equation and solve for ‘’X’’ ( 960 = 30x ) * The ration is 960 miles = x number of gallons and you put the “30” in because its miles per gallon.*

• When you have the equation “960 = 30x” divide 30 on both sides to isolate “x”.

Back to Problem

Page 123: CST Prep Part II MAIN MENU

Solution

960 = 30x 30 30 x = 32

Answer: A

Back to Problem Standard 15

Page 124: CST Prep Part II MAIN MENU

Standard 1616. Students understand the concepts of a relation and a function, determine

whether a given relation defines a function, and give pertinent information about given relations and functions.

Problem 92

Problem 93

Main Menu

Page 125: CST Prep Part II MAIN MENU

92) Beth is two years older than Julio. Gerald is twice as old as Beth.

Debra is twice as old as Gerald. The sum of their ages is 38. How old is Beth?

A 3B 5C 6D 8

Vocabulary Rules& Strategies

Solution & Answer Standard 16

Page 126: CST Prep Part II MAIN MENU

Vocabulary

• Twice - Double• Sum - Answer to an addition problem• Two years older - x +2

Back to Problem

Page 127: CST Prep Part II MAIN MENU

Rules & Strategies

Given: Find: Beth’s age• Beth is x• Julio is x-2• Gerald is 2x• Debra is 2(2x)• Total years of age is 38

Back to Problem

Add all of them to make the expression equivalent to 38

Page 128: CST Prep Part II MAIN MENU

Solution

x+(x-2) + 2x + 2(2x) = 38x + x - 2 + 2x + 4x = 38 +2 +28x = 408x = 408 8x=5

Answer: B Back to Problem Standard 16

Beth is five years old.

Combine like terms

Page 129: CST Prep Part II MAIN MENU

Vocabulary Rules & Strategies

Solution & Answer Standard 16

A.Input Output

1 2

2 2

3 3

4 3

B.Input Output

2 6

2 5

6 4

6 3

C.Input Output

1 2

2 4

4 6

4 8

D.Input Output

0 1

0 2

1 3

1 4

93) Which relation is a function?

Page 130: CST Prep Part II MAIN MENU

Vocabulary

• Relation = a property that associates two quantities in a definite order, as equality or inequality

• Function = set of ordered pairs in which none of the first elements of the pairs appears twice

Back to Problem

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Rules & Strategies

• Look @ input (x values) and the numbers in the x value can not be repeated twice.

Back to Problem

Page 132: CST Prep Part II MAIN MENU

Solution

Answer: A Back to Problem Standard 16

Input Output

1 2

2 2

3 3

4 3

Page 133: CST Prep Part II MAIN MENU

Standard 17 Students determine the domain of independent variables and the range of dependent variables

defined by a graph, a set of ordered pairs, or a symbolic expression.

Problem 94 Problem 95 Main Menu

Page 134: CST Prep Part II MAIN MENU

94) For which equation graphed below are all the y-values negative?

Vocabulary Rules and Strategies

Solution and Answer

Standard 17

a. b. c. d.

Page 135: CST Prep Part II MAIN MENU

Vocabulary

• y-value ; (x, y)

Back to Problem

Page 136: CST Prep Part II MAIN MENU

Rules and Strategies

• Lines must cross at negative y-int. (Quadrant 2 & 3) in order for all to be negative.

Back to Problem

Page 137: CST Prep Part II MAIN MENU

Solution and Answer

Lines crossin the y-intof -1.

Answer: AStandard 17 Back to

Problem

14

23

Page 138: CST Prep Part II MAIN MENU

95)What is the domain of the function shown on the graph below?

A {-1,-2,-3,-4}B {-1,-2,-4,-5}C {1,2,3,4}D {1,2,4,5}

Vocabulary Rules & Strategies

Solution & Answer Standard 17

Page 139: CST Prep Part II MAIN MENU

Vocabulary

• Domain=the set of all first coordinates (x-values) of a relation or function

• Function= A relation in which every domain value is paired with exactly one range value

Back to Problem

Page 140: CST Prep Part II MAIN MENU

Rules & Strategies

• Identify the ordered pairs (x,y)• Put them in a t –chart• Remember domain= x-values

Back to Problem

Page 141: CST Prep Part II MAIN MENU

Solution

Back to Problem Standard 17Answer: D; {1,2,4,5}

X y1 -1

2 -2

4 -4

5 -5

*Only focus on domain

Page 142: CST Prep Part II MAIN MENU

Standard 1818. Students determine whether a relation defined by a graph, a set of

ordered pairs, or symbolic expression is a function and justify the conclusion.

Problem 96Problem 96Problem 96Problem 96 Main Menu

Page 143: CST Prep Part II MAIN MENU

96) 96) Which of the following graphs represents a relation that is not a function of x?

Click here to look at graphs.VocabularyVocabularyVocabularyVocabulary

Solution and Solution and answeranswer

Solution and Solution and answeranswer

Rules and Rules and SrategiesSrategiesRules and Rules and SrategiesSrategies

Standard 18 Standard 18 Standard 18 Standard 18

Page 144: CST Prep Part II MAIN MENU

A A CC

BB D D

Back to Back to problemproblem Back to Back to

problemproblem

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vocabularyvocabulary1.1. Function_Function_ : A relation ship or expression

involving one or more variables. (y=mx+b)

2.2. Vertical line test: Vertical line test: putting strait lines going through X Axis.

It tests to see

Whether it is a

function or

not. Back to Back to problemproblem Back to Back to

problemproblem

Page 146: CST Prep Part II MAIN MENU

Solution Solution and and

answeranswerRemember

To Give each

a vertical line test D, isn`t a function Because it is being

hit more than once.

Back to Back to problemproblem Back to Back to problemproblem

Standard Standard 1818

Standard Standard 1818

Page 147: CST Prep Part II MAIN MENU

Rules and Strategies Rules and Strategies You have to give it the vertical line test in order to see if it is a function.

Rule: it is only a function if it only hits the line once.

Back to Back to problemproblem Back to Back to

problemproblem

Page 148: CST Prep Part II MAIN MENU

Standard 19 Students know the quadratic formula and are familiar with its

proof by completing the square.

Problem 64

Problem 63

Main Menu

Page 149: CST Prep Part II MAIN MENU

64) Four steps to derive the quadratic formula are shown below. What is the correct order for these

steps?

A) I, IV, II, IIIB) I, III, IV, IIC) II, IV, I, IIID) II, III, I, IV

Vocabulary Rules & Strategies

Solution & Answer Standard 19

Page 150: CST Prep Part II MAIN MENU

Vocabulary

Derive= to reach or obtain by reasoning; deduce; infer. Quadratic Formula= the formula for determining the

roots of a quadratic equation from its coefficients.Quadratic= involving the square and no higher power

of the unknown quantity; of the second degree. Formula= a general relationship, principle, or rule

stated, often as an equation, in the form of symbols.

Back to Problem

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Rules & Strategies Quadratic Proof

This is the

order a quadratic

formula proof should look like.

Back to Problem

Page 152: CST Prep Part II MAIN MENU

Solution

Solution: A

Back to Problem Standard 19

Page 153: CST Prep Part II MAIN MENU

#63Toni is solving this equation by completing the square.Step1: ax2 + bx =-cStep 2: x2 + b

a X= - ca

Step 3:?

Which should be step 3 in the solution?

A)X2 =-cb - b

aX B)X +ba = - c

ax

C)X 2 + baX+b

2a = -ca + b

2a D)X 2 +baX+ b

2a 2 = - c

a + b2a

2

VocabularyVocabulary Rules and SrategiesRules and Srategies

Solution and answer

Solution and answer Standard 19 Standard 19

Page 154: CST Prep Part II MAIN MENU

Vocabulary1.1. Completing the square: Completing the square: An expression in the

form of x 2 + bx is not a perfect square. However, you can use the relationship above to add a term to x 2 + bx to form a trinomial that can be a perfect square.

Back to problemBack to problem

Page 155: CST Prep Part II MAIN MENU

Solution and AnswerStep1: ax2 + bx =-c b/a 2Step 2: x2 + b

a X= - ca 2

Step 3:? ba x 1

2 2= b2a 2

x2 + ba X+ b

2a 2= - ca +now add to both

b2a 2 sides.

The answer is D

Back to problemBack to problem

1) Do completi

ng the square

1) Now add to both

sides2) Now you

have your answer.

Standard 19 Standard 19

Page 156: CST Prep Part II MAIN MENU

Rules and Strategies•First of you will need to divide b/a by 2 then you have to find the square root. Then add or subtract that number to both sides.• You’re completing the square.

Back to problemBack to problem

Page 157: CST Prep Part II MAIN MENU

Standard 20 Students use the quadratic formula to find the roots of a second-

degree polynomial and to solve quadratic equations.

Problem 65

Problem 67

Problem 66

Problem 68 Main Menu

Page 158: CST Prep Part II MAIN MENU

65) which is one of the solutions to the equation 2x2-x-4=0 ?

A. 1-√33 4

B. -1 +√33 4

C. 1+√33 4

D. -1-√33 4

Vocabulary Rules & Strategies

Solution & Answer Standard 20

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Vocabulary

• Solutions: the answer itself• Equation: A mathematical statement that two

expressions are equal

Back to Problem

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Rules & Strategies

• Identify a, b, and c • Use the quadratic formula -b+ √b2-4ac to solve

this equation 2(a)

Back to Problem

Page 161: CST Prep Part II MAIN MENU

Solution

2x2-x-4 A=2 -b + √b2-4ac 1-4(-8)B=-1 2(a) 1+32C=-4 -b+ √(-1)-4(20(4) 2(2) 33 1+√33 4 Answer : C

1+√33 1-√33 4 4 Back to

Problem Standard 20

Page 162: CST Prep Part II MAIN MENU

66.)Which statements best explains why there is no real solution to the quadratic

equation 2x2 + x + 7 = 0?

Vocabulary Rules and Strategies

Solution and Answer Standard 20

Page 163: CST Prep Part II MAIN MENU

Vocabulary

• Statement ; declaration of speech setting forth facts, particulars, etc.

• Solution ; an explanation or answer• Quadratic Equation ; ax 2 + bx + c = 0

Back to Problem

Page 164: CST Prep Part II MAIN MENU

Rules and Strategies

• Identify a, b , and c in ax 2 + bx + c• Find the discriminant : b 2 – 4ac• X>0 ; 2 solutions• X=0 ; 1 solution• X<0 ; no solution

Back to Problem

Page 165: CST Prep Part II MAIN MENU

Solution and Answer 2x 2 + x + 7 = 0Identify:a= 2b= 1c= 7Discriminant:b 2 – 4ac(1)2 – 4 2 7∙ ∙ =1 – 56 = - 55Negative!

Answer:C Back to Problem Standard 20

-55 < 0 ;No Solution

Page 166: CST Prep Part II MAIN MENU

67) What is the solution set of the quadratic equation 8x2+2x+1=0

A) -1 1 2, 4

B){-1+√2,-1 √2 }C) -1+√7, -1-√7

8 8D)No real solution

Vocabulary Rules & Strategies

Solution & Answer Standard 20

Page 167: CST Prep Part II MAIN MENU

Vocabulary

• Solution: the process of determining the answer to a problem

• Quadratic Equation: an equation containing a single variable of degree 2. Its general form is ax 2 + bx + c = 0, where x is the variable and a, b, and c are constants ( a ≠ 0).

Back to Problem

Page 168: CST Prep Part II MAIN MENU

Rules & Strategies

• Quadratic equation = ax2+bx+c=0• Plug in the values into the Quadratic equation• Simplify if possible• Answer

Back to Problem

Page 169: CST Prep Part II MAIN MENU

Solution

X=-b±√b2-4ac 2(a)

x=-2±√4-32 16 x=-2±√-28 Negative

16 = ΦAnswer: D

Back to Problem Standard 20

Page 170: CST Prep Part II MAIN MENU

68). What are the solutions to the equation 3x + 3 = 7x

A 7+ 85 or 7 - 85

B -7+ 85 or -7 - 85

C 7 + 13 or 7 - 13

D -7 + 13 or -7 - 13

Vocabulary Rules & Strategies

Solution & Answer Standard 20

2

6 6

6 6

6 6

6 6

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Vocabulary

• Solution = answer

• Inequality = a mathematical statement that two expressions are equal.

Back to Problem

Page 172: CST Prep Part II MAIN MENU

Rules & Strategies

Back to Problem

+ 2

Page 173: CST Prep Part II MAIN MENU

Solution 3x + 3 = 7x Inverse operation to get in correct form

-7x3x – 7x + 3 = 0 a=3, b=-7, c=3

7 (-7) – 4(3)(3)2(3)

7 136

Answer: C Back to Problem Standard 20

=x

+

+

2

2

2

Page 174: CST Prep Part II MAIN MENU

Standard 21 Students graph quadratic functions and know that their roots

are the x-intercepts.

Problem 69 Problem 71Problem 70

Problem 72 Main Menu

Page 175: CST Prep Part II MAIN MENU

69) The graph of the equation y=x²-3x-4 is shown below:

For what values or value of x is y=0?A x=-1 onlyB x=-4 onlyC x=-1 and x=4D x=1 and x=-4

Vocabulary Rules & Strategies

Solution & Answer Standard 21

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Vocabulary

• Equation= two equally balanced expressions

• Value= quantity

Back to Problem

Page 177: CST Prep Part II MAIN MENU

Rules & Strategies

• Identify x-intercept(s), root(s), zero(s), solution(s)

Back to Problem

Page 178: CST Prep Part II MAIN MENU

Solution

Solution:-1,4 Make sure to go

from left to right. Answer: C

Back to Problem Standard 21

x-intercept, root, solution, zeros x-intercept, root,

solution, zeros

Page 179: CST Prep Part II MAIN MENU

70) Which best represents the graph of y= -x²+3?

a. b.

c. d.

VocabularyVocabulary Rules and StrategiesRules and Strategies

SolutionSolutionStandard

21Standard

21

Page 180: CST Prep Part II MAIN MENU

Vocabulary

• Parabola = graph dervived from a quadratic function.

Back to ProblemBack to Problem

Page 181: CST Prep Part II MAIN MENU

Rules and Strategies

• After each step, make sure you graph before moving onto the next one.

• Follow the steps: 1)Axis of Symmetry2)Vertex3)Y-Intercept4)Two Other Points

Back to ProblemBack to Problem

Page 182: CST Prep Part II MAIN MENU

Solution and AnswerAxis of Symmetry = 0 Vertex = (0,3) Y intercept= y = -x²+3B=0 -(0) y= -(0)²+3 y= 3A= -1 2(-1) y=3

2 Other Points x= -1 x=-2Y= -(-1)²+3 y = -(-2)²+3Y= -1+3 y=-4+3y= 2 y= -1

Answer : B

Standard 21

Standard 21

Back to ProblemBack to Problem

Page 183: CST Prep Part II MAIN MENU

71) Which quadratic function when graphed has x-intercepts of

4 and -3?

A y= (x-3)(x+4) B y= (x+3) (2x-8)

C y= (3x-1) (3x+1) D y= (3x+1) (8x-2)

Vocabulary Rules & Strategies

Solution & Answer Standard 21

Page 184: CST Prep Part II MAIN MENU

Vocabulary

• X-intercepts: x-coordinates of the point where a graph intersects the x-axis.

• Quadratic function: A function that can be written in the form of f(x)= ax+bx=c where a, b, and c are real numbers and a zero.

Back to Problem

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Rules & Strategies

• Do zero product property.• Check your work.

Back to Problem

Page 186: CST Prep Part II MAIN MENU

Solutiony= (x-3)(x+4) y= (x+3)(2x-8) (x-3) (x+4)=0 (x+3)(2x-8)=0 x-3=0 x+4=0 x+3=0 2x-8=0 +3 +3 -4 -4 -3 -3 +8 +8 x=3 x=-4 x=-3 2x=8 3,-4 -3,4 2 2 x=-3 x=4 -3,4 = -3,4

Answer: BBack to Problem Standard 21

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72) What are the real roots of the function in the graph?:

A 3B -6C -1 and 3D -6, -1, and 3

Vocabulary Rules & Strategies

Solution & Answer Standard 21

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Vocabulary

• Roots: solution to function

Back to Problem

Page 189: CST Prep Part II MAIN MENU

Rules & Strategies

• Look at the x-axis!• Identify x-intercept(s), root(s), zero(s),

solution(s)

Back to Problem

Page 190: CST Prep Part II MAIN MENU

Solution

Solution:-1,3

Answer: c Back to Problem Standard 21

x-intercept, root, solution, zeros x-intercept, root,

solution, zeros

Page 191: CST Prep Part II MAIN MENU

Standard 22 Students use the quadratic formula or factoring techniques or both to

determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points.

Problem 73 Main Menu

Page 192: CST Prep Part II MAIN MENU

73) How many times does the graph of y = 2x2 – 2x + 3 intersect the x-axis?

a. Noneb. Onec. Twod. Three

Vocabulary Rules and Strategies

Solution and Answer

Standard 22

Page 193: CST Prep Part II MAIN MENU

Vocabulary

• x-axis - the horizontal axis in a two-dimensional coordinate system

• Discriminant- the name given to the expression that appears under the square root (radical) sign in the quadratic formula.

Back to Problem

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Rules and Strategies

• Try Super Diamond • Try discriminant

Back to Problem

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Standard 23 Students apply quadratic equations to

physical problems, such as the motion of an object under the force of gravity.

Problem 74 Problem 75 Problem 76 Main Menu

Page 197: CST Prep Part II MAIN MENU

74) An object that is projected straight downward with initial velocity v feet per second travels a distance s=vt+16t2 , where t= time in seconds.

If Ramon is standing on a balcony 84 feet above the ground and throws a penny straight down with initial velocity of 10 feet per second, in how many seconds will it reach the ground.

Vocabulary Rules & Strategies

Solution & Answer Standard 23

A) 2 secondsB) 3 secondsC) 6 secondsD) 8 seconds

Page 198: CST Prep Part II MAIN MENU

Rules & Strategies

• Plug in 10 feet per second under v.• Plug in 84 as s.

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Vocabulary

• Velocity=the rate of speed with which something happens; rapidity of action or reaction.

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SolutionS=vt+16t 2

84=10t+16t 2 -84 -84

0=16t 2+10t-84 Gcf:2

2 2 2 2x2x2x2x3x7

8(-42) 0=2(8t 2 +5t-42) 8x 8x 0=2(t-2)(8t+21) -16 21 t-2=0 or 8t+21=0 5 t=2 or t= -21

8

You have to pick the one that’s most Logical. The answer is A)2. Back to

Problem Standard 23

1)Substitute known values into equation.2)Put it in standard form.3)Solve by doing gcf and then super diamond.4)Now do zero property.5) Find out which one makes more sense because time can`t be expressed as a negative it has to be 2 seconds.

1)Substitute known values into equation.2)Put it in standard form.3)Solve by doing gcf and then super diamond.4)Now do zero property.5) Find out which one makes more sense because time can`t be expressed as a negative it has to be 2 seconds.

Page 201: CST Prep Part II MAIN MENU

75)The height of a triangle is 4 inches greater than twice its base. The area of the triangle is

168 square inches. What is the base of the triangle?

A) 7 in.B) 8 in.C) 12 in.D) 14 in.

Vocabulary Rules & Strategies

Solution & Answer Standard 23

Page 202: CST Prep Part II MAIN MENU

Vocabulary• Height-The perpendicular distance from any vertex of a triangle to the side

opposite that vertex. Also called altitude. Sometimes the height is determined OUTSIDE of the triangle.

• Base:The side of a triangle to which an altitude is drawn. the base and the altitude will be used to find the area

• area-A = 1/2(bh), where b is the length of the base, and h is the length of the altitude. A = Square root [s(sa)(sb)(sc)], where s is the semiperimeter and a, b, and c are the lengths of the sides of the triangle.

• triangle-a three-sided polygon • Altitude-segment from the vertex of a triangle perpendicular to the line containing

the opposite side.• Vertex-the point of intersection of lines or the point opposite the base of a figure

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Rules & Strategies

• Find the height of the triangle• Plug in area for area formula• solve the formula• Use zero product property• FACTOR by GCF then Diamond !

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Solution

Back to Problem Standard 23

X+2 A=168

b

12 168

14

168=1\2b(2b+4)

2(168)=2b2+4b

0=2b2+4b-336

x

+

-168

2

b b

-12 +14

0=2(b-12)(b+14)

B-12=0B=+12

B+14=0 B=-14

answer: C

Step:1 Step: 2

Step: 3

Step: 4

Step: 5

Step: 6

Page 205: CST Prep Part II MAIN MENU

76) A rectangle has a diagonal that measures 10 centimeters and a length that is 2 centimeters longer than the width. What is the width of the rectangle in centimeters?

a. 5b. 6c. 8d. 12

Vocabulary Rules and Strategies

Solution and Answer

Standard 23

Page 206: CST Prep Part II MAIN MENU

Vocabulary

• Diagonal - a line joining two nonconsecutive vertices of a polygon or polyhedron

• Length – The measurement of the extent of something from the vertical side

• Width - The measurement of the extent of something from side to side

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Rules and Strategies

1. Draw a diagonal in the rectangle2. Use Pythagorean's theory to solve for the

width.A. a=2+w, w=widthB. b= wC. c=diagonal, 10cm

3. (2+w)2+w2= 102

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Visual Picture

w

2+w10

Page 208: CST Prep Part II MAIN MENU

Solution (2+w)2+w2 = 102

(2+w)(2+w)+w2=100 4+4w+w2+w2=100 Combine like terms and put in descending -100 -100 order. 2w2+ 4w + -96 = Use super diamond method -192 w= 2 w 2 w =w 8 16 -12 -6 4 Answer: B

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(w+8)(w-6)= 0

Width = -8 or 6 width = 6

Standard 23

Page 209: CST Prep Part II MAIN MENU

Standard 25.1 Students use properties of numbers to construct simple, valid

arguments (direct and indirect) for, or formulate counterexamples to, claimed assertions.

Problem 23 Main Menu

Page 210: CST Prep Part II MAIN MENU

23) John’s solution to an equation is shown below. Given: x+5x+6=0

Step 1: (x+2)(x+3)=0 Step 2: x+2=0 or x+3=0

Step 3: x= -2 or x= -3Which property of real numbers did john use for Step 2?

A multiplication property of equality. B zero product property of multiplication.

C commutative property of multiplication. D distributive property of multiplication over

addition.

Vocabulary Rules & Strategies

Solution & Answer Standard 25.1

Page 211: CST Prep Part II MAIN MENU

Vocabulary

• Solution: the process of determining theanswer to a problem. • Equation: A mathematical statement that two

expressions are equal.• Zero Product Property: For real numbers p

and q, pq = 0 , then p= 0 or q =0 .

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Rules & Strategies

• Look at Step 2 and see which property it is.• Remember which property is which and don’t

mix them up.

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Solution

Step 2: x+2=0 or x+3=0Zero product property of

multiplication

Answer: BBack to Problem Standard 25.1

Page 214: CST Prep Part II MAIN MENU

52) What is the perimeter of the figure shown below, which is not drawn to scale?

Vocabulary Rules & Strategies

Solution & Answer Standard 10

X+13

3x

8

X+5

2

3x+2

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Vocabulary

• Perimeter: sum of all sides• Scale: size of the shape

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Rules and Strategies

• Add ALL of the sides • Combine like terms

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Solution and Answer

• 3x+2 + x+ 13+3x+8+2+x+5

• 8x + 30

• Answer: C

Back to Problem Standard 10

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77) What is x2 – 4xy+ 4 y2 reduced to lowest terms? 3xy-6y2

A)x-2y C) x+2y 3 3B) x-2y D) x+2y 3y 3y

Vocabulary Rules & Strategies

Solution & Answer Standard 12

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• Reduced: in simplest form

Vocabulary

Back to Problem

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• Look at numerator and decide which factoring method is needed.

• Look at denominator and decide which factoring method is needed.

• Divide out common factors.

Rules and Strategies

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x2 – 4xy+ 4 y2 (diamond) +4 3xy-6y2 ( GCF) x x (x-2y)(x-2y) divide out!! -2 y -4 -2y 3y (x-2y) (x-2y) Answer: B 3xy -6y2

3y 3y 3y 3y (x-2y)

Solution and Answer

Back to Problem Standard 12