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help2013 Group Title

solve inequality

  • one year ago
  • one year ago

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  1. help2013 Group Title
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    im writing info

    • one year ago
  2. help2013 Group Title
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    \[x-4=\sqrt{x+8}\]

    • one year ago
  3. bii17 Group Title
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    x will be equal to 1

    • one year ago
  4. help2013 Group Title
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    can u show me how plzzzzz! an dthanks

    • one year ago
  5. bii17 Group Title
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    \[(x-4)^2 = x+8\] \[x^2 -8x +16 = x+8\] combine similar terms it will become \[x^2 -9x+8 =0\] factor (x+9) (x-1) = 0 to check, substitute x=-9 or x=1

    • one year ago
  6. help2013 Group Title
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    thanks:)

    • one year ago
  7. tkhunny Group Title
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    \(x-4 = \sqrt{x+8}\) First, before ANYTHING else, think about the Domain. \(x+8\ge 0\;or\;x \ge -8\) You tell me why? \(x-4\ge 0\;or\;x \ge 4\) You tell me why? After that, if we get ANY result for x that is less than 4, we will simply discard it as extraneous. It should not even be considered as a solution. x = 1 is not correct. x = -9 is not correct.

    • one year ago
  8. help2013 Group Title
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    yeah thanks:)

    • one year ago
  9. tkhunny Group Title
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    Seriously. Go try them in the original equation. They fail miserably.

    • one year ago
  10. help2013 Group Title
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    well then is the answer zero

    • one year ago
  11. tkhunny Group Title
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    No, x = 0 is also incorrect. The only suspected results are x = 1 and x = 8. (The factoring shown above was incorrect.) x = 1 is immediately discarded as not in the Domain. x = 8 is a solution. x = -9 could not have been a solutions, as -9 < 1. x = 0 could not be a solution, as 0 < 1 If you rule out ALL your suspects, there is NO solution. In this case x = 8 is the winner.

    • one year ago
  12. help2013 Group Title
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    ohh well thats not one of my answer choices?:/

    • one year ago
  13. help2013 Group Title
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    opps nv thanks for the help

    • one year ago
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