C880 ALGEBRA FOR SECONDARY MATHEMATICS TEACHING —OUM1 Task 1 Assessing Algebra Concepts
Western Governors University
Western Governors University
ALGEBRA FOR SECONDARY MATHEMATICS TEACHING — C880
OUM1 TASK 1: ASSESSING ALGEBRA CONCEPTS
Sheila Pickrell
M.A. Mathematics Education (Middle Grades)
Explanation
Question Number Solution to the Problem Feedback for Student
of
Misconception
The error in the Great job on using
√𝑋2 − 4 = 5 student’s work is inverse operations when
1. √𝑋2 − 4 = 5 (√𝑋2 − 4)2 = (5)2 observed on the first solving the equation. I
step (highlighted). would like for you to
𝑋−2=5
𝑋2 − 4 = 25 The student thought
think about how to
2
𝑋 − 4 + 4 = 25 + 4 that getting the
𝑋 =5+2 square root of each remove the radical sign
𝑋2 = 29 term in an from the original
expression is the equation properly. What
𝑋=7 √𝑋2 = √29 same as taking the we have in your work
square root of the right now suggests that:
𝑋 = ±√29 whole expression. √𝑋2 − 4 = √𝑋2 − √4 = 𝑋 − 2
Student thinks that: Is this statement
√𝑋2 − √4 = √𝑋2 − 4 true? Remember that
once this is corrected,
there will be some
changes in the rest of
your solution.
(𝑋 − 2)2 = 21 The student Great job on
2. (𝑋 − 2)2 = 21 thought that the remembering how to
√(𝑋 − 2)2 = √21 parentheses use the distributive
𝑋 − 2 = ±√21 allows for the property. However, in
𝑋2 − 22 = 21 exponent “2” to this problem, you would
𝑋 − 2 + 2 = ±√21 + 2 be distributed to have to get rid of the
𝑋2 − 4 = 21
𝑋 = 2 ± √21 the “X” and the exponent “2” on the left
𝑋2 = 25 negative “2” side of the equation
(highlighted). instead of distributing it
𝑋=5 to “x” and “-2”. Try to
use what we have
learned to be the
opposite of the
exponent 2 to get rid of
it, and make sure to
keep equation balanced
at all
times.
Western Governors University
Western Governors University
ALGEBRA FOR SECONDARY MATHEMATICS TEACHING — C880
OUM1 TASK 1: ASSESSING ALGEBRA CONCEPTS
Sheila Pickrell
M.A. Mathematics Education (Middle Grades)
Explanation
Question Number Solution to the Problem Feedback for Student
of
Misconception
The error in the Great job on using
√𝑋2 − 4 = 5 student’s work is inverse operations when
1. √𝑋2 − 4 = 5 (√𝑋2 − 4)2 = (5)2 observed on the first solving the equation. I
step (highlighted). would like for you to
𝑋−2=5
𝑋2 − 4 = 25 The student thought
think about how to
2
𝑋 − 4 + 4 = 25 + 4 that getting the
𝑋 =5+2 square root of each remove the radical sign
𝑋2 = 29 term in an from the original
expression is the equation properly. What
𝑋=7 √𝑋2 = √29 same as taking the we have in your work
square root of the right now suggests that:
𝑋 = ±√29 whole expression. √𝑋2 − 4 = √𝑋2 − √4 = 𝑋 − 2
Student thinks that: Is this statement
√𝑋2 − √4 = √𝑋2 − 4 true? Remember that
once this is corrected,
there will be some
changes in the rest of
your solution.
(𝑋 − 2)2 = 21 The student Great job on
2. (𝑋 − 2)2 = 21 thought that the remembering how to
√(𝑋 − 2)2 = √21 parentheses use the distributive
𝑋 − 2 = ±√21 allows for the property. However, in
𝑋2 − 22 = 21 exponent “2” to this problem, you would
𝑋 − 2 + 2 = ±√21 + 2 be distributed to have to get rid of the
𝑋2 − 4 = 21
𝑋 = 2 ± √21 the “X” and the exponent “2” on the left
𝑋2 = 25 negative “2” side of the equation
(highlighted). instead of distributing it
𝑋=5 to “x” and “-2”. Try to
use what we have
learned to be the
opposite of the
exponent 2 to get rid of
it, and make sure to
keep equation balanced
at all
times.
