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sugarheart

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  • one year ago
  • one year ago

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  1. Mertsj
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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.

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

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

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

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

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

    • one year ago
  7. sugarheart
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    Yes

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

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

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

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

    • one year ago
  12. sugarheart
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    What?

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

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

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

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

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

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

    • one year ago
  19. precal
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    because 13 is extraneous

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

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

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

    • one year ago
  23. Mertsj
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    np

    • one year ago
  24. precal
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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

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

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

    • one year ago
  27. sugarheart
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    Yes lol

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

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

    • one year ago
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