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Goose.

  • one year ago

Find the solution of 4 times the square root of the quantity of x plus 2 equals negative 16, and determine if it is an extraneous solution. x = 14; not extraneous x = 2; extraneous x = 2; not extraneous x = 14; extraneous

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  1. Goose.
    • one year ago
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    \[4\sqrt{x+2}=-16\]

  2. anonymous
    • one year ago
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    any ideas on how to start?

  3. Goose.
    • one year ago
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    You would move -16 to the left side, and set it equal to 0?

  4. anonymous
    • one year ago
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    I'd divide by the 4.

  5. anonymous
    • one year ago
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    It's not a polynomial, so you don't have to set it equal to 0

  6. Goose.
    • one year ago
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    So I'd have? \[\frac{ \sqrt{x+2} }{ 4 }=-4\]

  7. anonymous
    • one year ago
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    |dw:1441897429536:dw|

  8. Goose.
    • one year ago
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    OH yeah, the 4 cancel eachother out on the left side,

  9. anonymous
    • one year ago
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    ok. so the everything left on the left side is under the radical, and the number on the right side is negative, so this equation has no real solution. And the solution must be extraneous

  10. anonymous
    • one year ago
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    because you can't take a square root and get a negative number as the result

  11. Goose.
    • one year ago
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    Okay,

  12. anonymous
    • one year ago
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    next is to square both sides

  13. Goose.
    • one year ago
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    So x + 2 = 16

  14. anonymous
    • one year ago
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    right

  15. Goose.
    • one year ago
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    Then you isolate your variable, to make it x = 14, extraneous.

  16. Goose.
    • one year ago
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    Thanks :)

  17. anonymous
    • one year ago
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    you're welcome

  18. radar
    • one year ago
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    What would be the results, if you divided both sides by a -4?|dw:1441898080802:dw|

  19. radar
    • one year ago
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    Squaring both sides getting: x+2 = 16 x=14 \[4\sqrt{14 + 2}=-16\]\[4\sqrt{16}=-16\]\[4\times \pm4=\pm16\]\[\pm16=\pm16\] If we accept that the square root of 16 is a plus or minus 4, then it would seem that 14 is not extraneous Hmmmmm

  20. radar
    • one year ago
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    With that I will log off and do some review.....

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