Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 1.a) Factor:...

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

1. a) Factor: x2 + 10x + 25 = 0

* Identify a, b and c. a = 1 , b = 10 , c = 25 * You are looking for two numbers that do the following:

Add up to give you b and multiply to give you c.

(x + 5)(x + 5) = 0 Or (x + 5)2 = 0

Notice: 5 + 5 = 10

b) Use the zero product property to find the

solutions.

and 5 • 5 = 25

(x + 5)(x + 5) = 0 Factored form

Zero Product Propertyx + 5 = 0 – 5– 5 Subtract

x = − 5© 2007- 09 by S-Squared, Inc. All Rights Reserved.

1. Factor: x2 + 10x + 25 = 0

c) Check your solution. x = − 5

x2 + 10x + 25 = 0 (− 5)2 + 10(− 5) + 25 = 0

25 – 50 + 25 = 0

− 25 + 25 = 0

0 = 0

Check

Equation

Simplify

Substitute


2. a) Factor: x2 – 5x – 24 = 0

* Identify a, b and c. a = 1, b = − 5 , c = − 24

* You are looking for two numbers that do the following:


Notice: 3 + (− 8) = − 5 and 3 • (− 8) = − 24


(x + 3)(x – 8) = 0

b) Use the zero product property to find the

solutions.(x + 3)(x – 8) = 0 Factored form

Zero Product Propertyx + 3 = 0 and x – 8 = 0– 3– 3Subtract

x = − 3

+ 8+ 8 Add

x = 8

3. a) Factor: x2 – 81 = 0

Difference of two perfect squares pattern.

* a2 – b2 = (a ─ b)(a + b)

* Identify the a and the b by taking the square root of each term.

Notice:

x2 x = a

9 81

= b

(x – 9)(x + 9) = 0

* Substitute into the difference of two perfect square pattern.


3. Factor: x2 – 81 = 0

(x – 9)(x + 9) = 0 Factored form

Zero Product Propertyx – 9 = 0 and x + 9 = 0

+ 9+ 9Add

x = 9

– 9– 9 Subtract

x = − 9


b) Use the zero product property to find thesolutions.

a) Factor out the common monomial.

4. Given: 4x2 – 14x + 6 = 0

2(2x2 – 7x + 3) = 0

* The greatest common monomial is 2.

Factor out a 2


b) Factor the resulting trinomial

4. Given: 4x2 – 14x + 6 = 0

2(2x2 – 7x + 3) = 0

−6

* Identify a, b and c. a = 2, b = − 7 , c = 3


Add up to give you b and multiply to give you a • c.

2(x – 3)(2x – 1) = 0

Notice: − 6 + (−1) = − 7 and − 6 • (− 1) = 6

2x 2x −1

and Reduce −3 x 2x

−1and

* Write your final factorization using the two fractions.

* Build fractions using the leading term less one degree as your numerator and −6 and −1 as your denominators.

1

−3


c) Use the zero product property to findthe solutions

4. Factor: 4x2 – 14x + 6 = 0

2(x – 3)(2x – 1) = 0 Factored form

Zero Product Propertyx – 3 = 0 and 2x – 1 = 0

+ 3+ 3Add

x = 3

+ 1+ 1 Add

1

Divide 2 2

2

2x = 1

x =


h = height of the object at time tt = time in secondss = initial height

5. Using the Vertical Motion Model h = − 16t2 + swhere

You are standing on a cliff 784 feet high and drop your cell phone. How long will it take until your phone hits the ground below (h = 0)?

h = 0t = t (need to find)s = 784

* Identify h, t and s

Substitute0 = − 16t2 + 784

– 784

– 784

Subtract

Isolate t2

− 784 = − 16t2

Divide −16 −16

49 = t2


h = height of the object at time tt = time in secondss = initial height

5. Using the Vertical Motion Model h = − 16t2 + swhere

You are standing on a cliff 784 feet high and drop your cell phone. How long will it take until your phone hits the ground below (h = 0)?

49 = t2+ Square root

7 = t+

* Time is always positive

7 = t

It will take 7 seconds for the phone to hit the ground


6. Complete the following given:

a) Let y = 0, factor and solve using the zero product property.

a = 1b = 2 c = − 8



0 = (x + 4)(x – 2)

Notice: 4 + (− 2) = 2 and 4 • (− 2) = − 8

Factored form

Zero Product Propertyx + 4 = 0 and x – 2 = 0– 4– 4Subtract

x = − 4

+ 2+ 2 Add

x = 2

0 = x2 – 6x + 8 Let y = 0


y = x2 + 2x – 8


b) Identify the x – intercepts.

a = 1b = 2 c = − 8

x = − 4 x = 2 From part (a)

* The zeros of the quadratic are also the x – intercepts.

(− 4, 0) and (2, 0)

* Notice, the x – intercepts have a y – coordinate of 0.


y = x2 + 2x – 8

− 22

y = x2 + 2x – 8


c) Find the vertex.

x =− 4 + 2

2

x = − 1

The x-value of vertex is the mid-point of thex-intercepts.

x-value of vertex

Simplify

* The vertex is the highest or lowest point on a parabola.


(− 4, 0) and (2, 0)

y = x2 + 2x – 8


c) Find the vertex.

x = − 1* Substitute into the quadratic equation to find y.

y = x2 + 2x – 8

y = (−1)2 + 2(− 1) – 8

y = 1 – 2 – 8

y = − 1 – 8

y = − 9 Vertex (−1, − 9)

Equation

Simplify

Substitute



d) Find the y – intercept.

y = x2 + 2x – 8

* Where the graph crosses the y-axis, NOTE: x = 0

(0, − 8)

y = x2 + 2x – 8

y = (0)2 + 2(0) – 8

y = 0 + 0 – 8

y = − 8


x – intercepts: (− 4, 0) and (2, 0)


e) Graph:

Vertex: (− 1, − 9)

* Plot the following ordered pairs:

y – intercept: (0, − 8)


y = x2 + 2x – 8

You are aMath Super Star

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 1.a) Factor:...

Documents

Transcript of Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 1.a) Factor:...