AnswersFL-1200330-Algebra 2 CRMultiplying Radicals

Radical Equations and Extraneous Roots Answers

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1
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Solve:

Question illustration
A
x = –4
B
x = 4
C
x = –4 or x = 4
D
no real solution
2
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Solve:

Question illustration
A
s = 2
B
s = 7
C
s = 2 or s = 7
D
no real solution
3

Solve:

Question illustration
A
x = 2
B
x = 10
C
x = 2 or x = 10
D
no real solution
4

Solve:

Question illustration
A
Option
Option A
B
a = 1
Option B
C
Option
Option C
D
no real solution
Option D
7

in.

Answer:

8

8

Do you agree? Explain why or why not.

Question illustration
Answer:

I do not agree with the student. While solving the equation results in x 4, this is an extraneous solution. When you substitute 4 back into the original equation, you get sqrt(2(4) 1) 3 0, which simplifies to 3 3 0, or 6 0. Since this is a false statement, x 4 is not a solution. Furthermore, the equation sqrt(2x 1) 3 has no real solution because a principal square root cannot be negative.

9

Did your explanation include the following?

A
I do not agree with the student because there is no real solution.
B
Solving the equation only gives possible solutions, but you must check them in the original equation.
C
Checking the solution in the original equation shows that 4 is an extraneous solution.
10

Do you agree with the student? Explain why or why not.

A
I do not agree with the student because there is no real solution.
Option A
B
Checking the solutions in the original equation verifies [image 1] ; however, [image 2] is extraneous.
Option B
C
Checking the solutions in the original equation shows that 6 is an extraneous solution.
Option C
D
I agree with the student because both [image 1] and [image 2] satisfy the original equation.
Option D

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Radical Equations and Extraneous Roots Answers…