
Solving Quadratic Equations Using Substitution

Interactive Video
•
Mathematics
•
8th - 12th Grade
•
Hard
+1
Standards-aligned

Aiden Montgomery
FREE Resource
Standards-aligned
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10 questions
Show all answers
1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the initial form of the equation given in the problem?
r^2 + 2r - 3 = 0
(r + 4)^2 + 2(r + 4) - 3 = 0
r^2 + 4r + 2 = 0
(r + 4)^2 - 3 = 0
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What substitution is made to simplify the equation?
u = r - 4
u = r^2 + 4
u = r + 4
u = 2r + 4
Tags
CCSS.HSA-REI.B.4B
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
After substitution, what is the form of the quadratic equation?
u^2 + 2u - 3 = 0
u^2 - 2u + 3 = 0
u^2 + 3u - 2 = 0
u^2 - 3u + 2 = 0
Tags
CCSS.HSA-REI.B.4B
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What are the factors of the quadratic equation in terms of u?
(u + 4)(u - 1)
(u + 2)(u - 2)
(u + 3)(u - 1)
(u + 1)(u - 3)
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the next step after factoring the equation in terms of u?
Solve for u
Substitute back to solve for r
Verify the solution
Graph the equation
Tags
CCSS.HSA.APR.A.1
6.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What are the simplified factors in terms of r?
(r + 5)(r + 2)
(r + 7)(r + 3)
(r + 6)(r + 4)
(r + 8)(r + 2)
Tags
CCSS.8.EE.C.8B
CCSS.HSA.REI.C.6
7.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What are the solutions for r after solving the linear equations?
r = -5 and r = -2
r = -6 and r = -4
r = -7 and r = -3
r = -8 and r = -2
Tags
CCSS.HSA-REI.B.4B
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