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magepker728

  • 2 years ago

x^2-6=0 why isnt it (x+2) (x-3)=0 just need to know why

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  1. hartnn
    • 2 years ago
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    because 3*3=9 and not 6

  2. hartnn
    • 2 years ago
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    x^2-9 = x^2-3^2 = (x+3)(x-3)

  3. hartnn
    • 2 years ago
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    x^2-6 = x^2-9+3 = (x+3)(x-3)+3

  4. magepker728
    • 2 years ago
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    typo its suppose to be (x-3) (x+2)

  5. hartnn
    • 2 years ago
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    ohh, (x-3)(x+2) = x^2 -3x+2x -6 =x^2 -x -6 and not x^2 -6

  6. magepker728
    • 2 years ago
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    if factored is it (x-3) (x+2) or square root +-6?

  7. hartnn
    • 2 years ago
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    roots are \(\large \pm \sqrt6\) factors are \((x+\sqrt6)(x-\sqrt6) \)

  8. magepker728
    • 2 years ago
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    but it is not (x-3) (x+2)

  9. hartnn
    • 2 years ago
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    exatly! its not because (x-3)(x+2) = x^2-x-6

  10. magepker728
    • 2 years ago
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    ooh i see

  11. akash809
    • 2 years ago
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    (x-3) (x-2)= x^2-2x-3x+6=x^2-5x+6...so u got an extra 5x

  12. hartnn
    • 2 years ago
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    ask if still any doubts.

  13. magepker728
    • 2 years ago
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    so when ever it is not a perfect square, i have to to take the square root of both side?

  14. CliffSedge
    • 2 years ago
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    You have to take the square root regardless.

  15. hartnn
    • 2 years ago
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    whenever ? when its in the form of x^2 - a^2 = 0

  16. hartnn
    • 2 years ago
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    then shift a^2 to other side and take square root.

  17. magepker728
    • 2 years ago
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    k thanks

  18. hartnn
    • 2 years ago
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    welcome :)

  19. febylailani
    • 2 years ago
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    bcs u hv 1 exttra x :)

  20. magepker728
    • 2 years ago
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    :D

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