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Solving Quadratic Equations Algebraically

Authored by Anthony Clark

Mathematics

10th Grade

CCSS covered

Solving Quadratic Equations Algebraically
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15 questions

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1.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Which of the following represents the first step in solving:

y = x2 + 3x - 5

y = x + 3

x2 - 3x + 5 = x - 3

x2 + 3x - 5 = x + 3

x2 + 3x - 5 + x + 3 = 0

x2 + 3x - 5 = 0

Tags

CCSS.HSA-REI.B.4B

2.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

What are the solutions to the system:
y = x2 + 3x - 5
y = x + 3

(-4, 2) and (-1, 5)

(-4, -1) and (2, 5)

(2, -1 and (-4, 5)

(-4, 1) and (2, -1)

Tags

CCSS.HSA.REI.C.7

3.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Find the solution:
y = x2 - 4x + 6
y = x + 2

(3, 1) and (6, 4)

(-1, 1) and (4, 2)

(1, 3) and (4, 6)

(1, 4) and (3, 6)

Tags

CCSS.HSA.REI.C.7

4.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Solve the system of linear and quadratic equations using substitution:
y = 5
y = 2x2 - 16x + 29

(6, 5) and (2, 5)

(2, 6)

(6, 2)

(5, 6) and (5, 2)

Tags

CCSS.HSA.REI.C.7

5.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Media Image

Determine the number of solutions to the given linear-quadratic system.

0 solutions

1 solution

2 solutions

3 solutions

Tags

CCSS.HSA.REI.C.7

6.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Solve by factoring: x2 + 3x - 18 = 0

x = -3 and x = 6

x = 3 and x = -6

x = -9 and x = 2

x = 9 and x = -2

Tags

CCSS.HSA-REI.B.4B

7.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Solve
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

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