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anonymous

  • one year ago

PLEASE HELP, IM LOST ON THIS QUESTION Given the equation 2Square root of x minus 5 = 2, solve for x and identify if it is an extraneous solution.

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  1. anonymous
    • one year ago
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    @@peachpi @imqwerty @arindameducationusc @godgirl122 @heretohelpalways please help I'm confused

  2. anonymous
    • one year ago
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    i did already but i got 2x-5=4

  3. anonymous
    • one year ago
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    or would it be 4x-5=4?

  4. imqwerty
    • one year ago
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    is this the question - \[\sqrt[2]{x-5}=2 \]??

  5. anonymous
    • one year ago
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    yes

  6. imqwerty
    • one year ago
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    ok so when u square both the sides \[(\sqrt[2]{x+5})^2 =2^2\] what will be the RHS after squaring? :)

  7. anonymous
    • one year ago
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    sorry , whats the RHS again?

  8. imqwerty
    • one year ago
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    RHS=right hand side :)

  9. anonymous
    • one year ago
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    ok thanks it would be

  10. imqwerty
    • one year ago
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    right hand side of the equation

  11. anonymous
    • one year ago
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    4

  12. imqwerty
    • one year ago
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    yes and what wuld be the Left hand side?

  13. anonymous
    • one year ago
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    x+5

  14. imqwerty
    • one year ago
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    so we get x+5 =4

  15. anonymous
    • one year ago
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    yes, now we solve for the variable and get -1 right?

  16. anonymous
    • one year ago
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    @imqwerty

  17. anonymous
    • one year ago
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    please help i don't know what to do after this step

  18. imqwerty
    • one year ago
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    yes u get x=-1 :)

  19. anonymous
    • one year ago
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    @imqwerty wait don't go please help

  20. imqwerty
    • one year ago
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    ok

  21. imqwerty
    • one year ago
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    u need help in??

  22. anonymous
    • one year ago
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    well my answer choices are these : x= 6, solution is not extraneous x = 6, solution is extraneous x = 11, solution is not extraneous x = 11, solution is extraneous

  23. anonymous
    • one year ago
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    and none of those are -1 so i don't know what to do next

  24. anonymous
    • one year ago
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    @imqwerty @peachpi

  25. imqwerty
    • one year ago
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    wait

  26. imqwerty
    • one year ago
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    we got x-5 = 1 add 5 to both sides u get x=6

  27. anonymous
    • one year ago
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    ok thank you so much

  28. anonymous
    • one year ago
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    how do i know if it is extraneous or not

  29. imqwerty
    • one year ago
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    and the solution is not extraneous cause when u put x=6 is a true solution :)

  30. anonymous
    • one year ago
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    ok so if its a true solution its always not extraneous?

  31. imqwerty
    • one year ago
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    yes :)

  32. anonymous
    • one year ago
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    ok, thank you for all of your help! it was much appreciated

  33. imqwerty
    • one year ago
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    no prblem :)

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