Determining Extraneous Roots (2.9.1) Flashcards

1
Q

• Extraneous roots are solutions that don’t work when you use them in the original equation.

A

• Extraneous roots are solutions that don’t work when you use them in the original equation.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

• Always check all solutions in the original equation so you can determine which ones work and which ones do not work.

A

• Always check all solutions in the original equation so you can determine which ones work and which ones do not work.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

• Always check your answer when you have a variable in the denominator because the denominator can never have a value of zero.

A

• Always check your answer when you have a variable in the denominator because the denominator can never have a value of zero.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

• Take only the principal (positive) square root when checking equations with radicals for extraneous roots.

A

• Take only the principal (positive) square root when checking equations with radicals for extraneous roots.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

This basic example worked out on the left is used to make the point that some solutions don’t work and are therefore what are known as extraneous roots.

Take what is given and square it. (This is not something you would ordinarily do, but we’re making a point here.)

Solve the resulting equation for x.

And, as usual, you get two answers.

Check both answers in the original equation and it is obvious that one of them, –2, doesn’t work.

Therefore, –2 is an extraneous root, not a solution

A

This basic example worked out on the left is used to make the point that some solutions don’t work and are therefore what are known as extraneous roots.

Take what is given and square it. (This is not something you would ordinarily do, but we’re making a point here.)

Solve the resulting equation for x.

And, as usual, you get two answers.

Check both answers in the original equation and it is obvious that one of them, –2, doesn’t work.

Therefore, –2 is an extraneous root, not a solution

How well did you know this?
1
Not at all
2
3
4
5
Perfectly