Practice Exam Quadratic Equations .docx

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Math 20-2: Quadratic Equations Practice Exam

Name: ___________________________ Period: _____________________ Date: ___________________________ Teacher: _____________________

Show your work in the space beside each question. Diagrams shown are not drawn to scale.

Multiple Choice

Identify the choice that best completes the statement or answers the question.

1. Solve x2_{– x – 6 = 0 by graphing the corresponding function and determining the zeros.}

a. x = 3, x = 2 b. x = 2, x = –3 c. x = –3, x = –2 d. x = 3, x = –2

2. Solve –2x2_{– 6x = –20 by graphing the expressions on both sides of the equation.}

a. x = 5, x = –2 b. x = –4, x = 5 c. x = 4, x = –5 d. x = –5, x = 2

3. Rewrite x2_{+ 2x = –3x}2₊_{2x + 36 in standard form. Then solve the equation in standard form by graphing.}

a. x = 6, x = 6 b. x = 3, x = 3 c. x = 6, x = –6 d. x = 3, x = –3

4. Solve x2_{+ 5x + 4 = 0 by factoring.}

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a.

x = , x = 4 b. x = –4, x = –3 c.

x = – , x = –4 d. x = 4, x = 3

6. Solve 10 + 5x2_{+ 18x = –4x}2_{– 18x – 10 by factoring.}

a.

x = , x = b.

x = – , x = – c.

x = , x = d.

x = – , x = –

7. Solve x2_{+ 6x + 5 = 0 using the quadratic formula.}

a. x = 5, x = –1 b. x = –5, x = –1 c. x = –5, x = 1 d. x = 5, x = 1

Use the following information to answer the next question

1 2 3 4 5 –1 –2 –3 –4 –5 x 1 2 3 –1 –2 –3 –4 –5 –6 –7 y

8. What is the correct quadratic function for the parabola? a. f(x) = (x + 2)(x + 3)

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9. Solve x2_{– 2x = 4 using the quadratic formula.}

a. _{x = 1 +} _{, x = 1 –} b. _{x = –1 +} _{, x = –1 –} c. _{x = –1 +} _{, x = –1 –} d. _{x = 1 +} _{, x = 1 –}

10. Solve 4x2_{– 2x = –3x}2_{+ 2 using the quadratic formula.}

a.

x = – , x = – b.

x = , x =

c.

x = – , x = – d.

x = , x =

11. A bridge is supported by three arches. The function that describes the arches is

h(x) = –0.2x2_{+ 3.0x, where h(x) is the height, in metres, of the arch above the ground at any distance, x,} in metres, from one end of the bridge. How tall is each arch?

a. 9.7 m b. 11.3 m c. 10.8 m d. 13.5 m

12. A hockey arena sells premium tickets for $70. At this price, the arena will sell 150 premium tickets every game. The owners know from past years that they will sell 3 fewer premium tickets per game for each price increase of $2. What should the owners charge for a premium ticket to earn the maximum amount of money?

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Short Answer

1. Determine the roots of the corresponding quadratic equation for the graph.

2. Solve 2y2_{+ 4y – 30 = 0 by factoring. Verify your solution.}

3. Solve . State the solution as exact values.

Problem

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2. a) Write a quadratic function with zeros at and and goes through (1, 2)

3. A theatre sells tickets to a musical. The profit function for the show is

p(t) = –30t2_{+ 550t – 400, where p(t) is the profit and c is the price of each ticket,} both in dollars.

a) What ticket price will result in the theatre breaking even on the show?

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summer. They usually sell 15 car washes per day. Experience has taught them that a $1 decrease in the price of a car wash results in 5 more sales per day. Let “x” equal to the number of times the price has decreased.

Number of $1 decreases, x

Price of car wash Number of cars washed Total Revenue

1

5

10

20

 Complete the above table

 State the Revenue equation (hint: Revenue = (Number of cars) (Price)) (1 mark)

R(x) = ( ) ( )

 Determine the two x-intercepts for the parabola and explain what they represent in this context (2 mark)

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practice exam Answer Section

MULTIPLE CHOICE

1. D 2. D 3. D 4. A 5. C 6. D 7. B 8. B 9. D 10. D 11. B 12. C

SHORT ANSWER

1. ANS: x = 1, x = –3

2. ANS: y = –5, y = 3

3. ANS:

,

PROBLEM

1. ANS:

Let l represent the length of the garden, in metres. Let w represent the length of the garden, in metres. w = l – 2

Let A represent the area of the garden, in square metres. lw = A

l(l – 2) = 36 l2_{– 2}_l_{– 36 = 0}

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The length is 7.083 m.

Therefore, the width is 5.083 m.

2. ANS:

a)

b) Other possible functions are multiples of the function where a 1. 

Examples may vary: f(x) = 2x2_{– 2.50}_x_{+ 0.650} f(x) = –2x2_{+ 2.50}_x_{– 0.650}

3. ANS:

a) 0 = –30t2_{+ 550}_t_{– 400}

Divide both sides by 10. a = –3, b = 55, c = –40

or

The price of a ticket must be $0.76 or $17.57 for the charity to break even.

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4.

Number of $1

decreases, x Price of car wash Number of cars washed Total Revenue

1 24 20 480

5 20 40 600

10 15 65 975

20 5 115 575

 Revenue equation: R(x) = (15 – 5x)(25 – x)  To find x – intercepts set R(x) = 0

Therefore the x-intercepts are -3 and 25

They represent the amount of decreases that would give you 0 revenue (no profit).

 METHOD 1:

Graph y1 = (15 – 5x)(25 – x) Find the vertex (2nd_{, trace, 4:max)}

Vertex (11, 980)

Thererfore max profit is $980

METHOD 2

Since the x-intercepts are -2 and 25, we can find the middle by averaging the intercepts

Therefore, 11 decreases Price = 25 – 11 = $14