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anonymous

  • one year ago

PLEASE HELP!!!

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @jim_thompson5910 @zzr0ck3r

  3. anonymous
    • one year ago
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    square both sides, what do you get?

  4. anonymous
    • one year ago
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    I got x = 10, but I'm not sure if that is an extraneous solution or not

  5. anonymous
    • one year ago
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    show your work, please

  6. anonymous
    • one year ago
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    |dw:1438221152504:dw|

  7. anonymous
    • one year ago
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    but when we plug x=10 into the original equation, we have 12=-12, right?

  8. anonymous
    • one year ago
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    right

  9. anonymous
    • one year ago
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    so?

  10. anonymous
    • one year ago
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    I don't know. @dan815

  11. anonymous
    • one year ago
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    What is the field? Real or Complex?

  12. anonymous
    • one year ago
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    Real

  13. anonymous
    • one year ago
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    Actually, with x =10 then \(\sqrt{16}=\pm 4\) and we just use -4 to get the right solution, but I don't know how to argue further if the field is real.

  14. anonymous
    • one year ago
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    And there is no extraneous solution

  15. Nnesha
    • one year ago
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    need help?

  16. Nnesha
    • one year ago
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    substitute x by its value if you get equal sides then 10 isn't an extraneous if both sides are not equal then 10 is an extraneous solution

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