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sugarheart

  • one year ago

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  1. Mertsj
    • one year ago
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    The deal is this....when you square an equation you sometimes end up with a "solution" that doesn't make the sentence true. Those are called extraneous roots. They are the reason your teacher tells you to always check the answers you get when solving radical equations. Sometimes they don't work. Let me show you an example.

  2. Mertsj
    • one year ago
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    |dw:1366677654241:dw|

  3. precal
    • one year ago
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    also, the domain helps in identifying extraneous solutions

  4. Mertsj
    • one year ago
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    And that answer will not check. It is an extraneous solution.

  5. sugarheart
    • one year ago
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    I got it, it's 5

  6. Mertsj
    • one year ago
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    You solved your problem?

  7. sugarheart
    • one year ago
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    Yes

  8. Mertsj
    • one year ago
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    Good for you!!!

  9. Mertsj
    • one year ago
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    Hey...just a minute. I don't think 5 will check.

  10. Mertsj
    • one year ago
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    |dw:1366677868347:dw|

  11. Mertsj
    • one year ago
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    Some graduate student made a grievous error in writing your question.

  12. sugarheart
    • one year ago
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    What?

  13. Mertsj
    • one year ago
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    As you can see...5 does not make the sentence true.

  14. Mertsj
    • one year ago
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    So if 5 was accepted as a correct answer, someone has put the wrong answer into the computer program .

  15. Mertsj
    • one year ago
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    Didn't you also get -3 as an answer?

  16. precal
    • one year ago
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    you mean |dw:1366678251238:dw|

  17. sugarheart
    • one year ago
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    @precal Why does that matter though.. It wasn't my question it was his example

  18. Mertsj
    • one year ago
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    |dw:1366678291957:dw|

  19. precal
    • one year ago
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    because 13 is extraneous

  20. Mertsj
    • one year ago
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    You see...-3 is actually an answer. It checks. 5 is extraneous.

  21. Mertsj
    • one year ago
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    @precal That was an example I constructed to illustrate extraneous roots for the asker

  22. precal
    • one year ago
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    sorry Mertsj, saw your example and not the actual problem

  23. Mertsj
    • one year ago
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    np

  24. precal
    • one year ago
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    Key idea @sugarheart is that when solving solutions for square root equations or radicals, extraneous solutions do exist so we have to check for them

  25. sugarheart
    • one year ago
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    5 is the extraneous solution and that's why it's was the answer to my problem

  26. Mertsj
    • one year ago
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    oh. So they were asking for the extraneous root. Cool. Then it is right.

  27. sugarheart
    • one year ago
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    Yes lol

  28. Mertsj
    • one year ago
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    Yep. I reread the question. Always helps to know the question one is attempting to answer.

  29. precal
    • one year ago
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    extraneous solutions can also, occur in other functions (example logarithmic)

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