Unit 1 Test

Identify the 'a' value: y = 16x² -8x -24

• 16
• 8
• -8
• -24

4ac would equal:  y = 2x² -x -3

• -12
• 8
• -24
• -1

Tags

CCSS.HSA-REI.B.4B

Solve     3x² - 11x + 10

• 1.7 or 2
• 1 or 2
• 1.5 or 2
• -1.7 or 2

Tags

CCSS.HSA-REI.B.4B

Solve     8x² - 6x + 1 = 0

• No Solution
• 0 or 1
• -0.5 or -0.25
• 0.25 or 0.5

Tags

CCSS.HSA-REI.B.4B

Solve    2x² + 7x - 15 = 0

• -1.5 or 5
• No Solution
• -5 or 1.5
• 0.7 or 5

Tags

CCSS.HSA-REI.B.4B

Solve using the quadratic formula:
4x² + 4x + 1 = 0

• x = -½
• x = -2
• x = 0

Tags

CCSS.HSA-REI.B.4B

Factor
k²+7x+10

• (k+2)(k+5)
• (k-2)(k-5)
• (k+10)(k+4)
• (k+2)(k-5)

Tags

CCSS.HSA.APR.A.1

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

What are the factors AND solutions of x² + 2x – 3 = 0

• (x - 2)(x + 1); x=2, x=-1
• (x + 1)(x - 3); x=-1, x=3
• (x + 2)(x - 1); x=-2, x=1
• (x - 1)(x + 3); x=1, x=-3

Tags

CCSS.HSA-REI.B.4B

What is the first step in solving 
(x + 2)(x - 3) = 0      ?

• Plug in zero for x
• Solve for x
• FOIL
• Set each binomial factor equal to zero

Tags

CCSS.HSA-REI.B.4B

Solve by factoring:

x² + 2x - 24 = 0

• x = -6, x = 4

• x = 6, x = -4

• x = 12, x = -2

• x = 8, x = -3

Tags

CCSS.HSA-REI.B.4B