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anonymous

  • 5 years ago

the algebraic expressions x-2 divided by x² -9is undefined when x is ..?

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  1. anonymous
    • 5 years ago
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    It's undefined when the denominator is 0. This occurs when \[x^2-9=0 \rightarrow x=\pm 3\]

  2. anonymous
    • 5 years ago
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    if the denominator is zero than wouldnt the answer be zero?

  3. anonymous
    • 5 years ago
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    No, the denominator is zero for x-values such that \[x^2-9=0\] By what you're saying, if the answer is zero, x=0. But then the denominator would be \[0^2-9=-9\]which is fine. You can divide your expression by -9.

  4. anonymous
    • 5 years ago
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    \[\frac{x-2}{x^2-9}=\frac{something}{0}\]is undefined.

  5. anonymous
    • 5 years ago
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    You have to find the x-values that make the denominator 0.

  6. anonymous
    • 5 years ago
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    Is that clearer?

  7. anonymous
    • 5 years ago
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    not really

  8. anonymous
    • 5 years ago
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    Is it the 'undefined' thing, i.e. *why* you have to find x where x^2-9 = 0?

  9. anonymous
    • 5 years ago
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    yea.. i dont understand that..

  10. anonymous
    • 5 years ago
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    Okay. In division, you can't ever divide by 0. This is because, if we did it, we wouldn't actually end up with something that made sense. I'll give you a simple example.

  11. anonymous
    • 5 years ago
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    ok

  12. anonymous
    • 5 years ago
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    You're happy with\[\frac{6}{2}=3\] yeah?

  13. anonymous
    • 5 years ago
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    yes

  14. anonymous
    • 5 years ago
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    So if I then write,\[6=\left[ ? \right]\times 3\]what number would you put in there?

  15. anonymous
    • 5 years ago
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    2

  16. anonymous
    • 5 years ago
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    Right. Now,

  17. anonymous
    • 5 years ago
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    if we imagine \[\frac{6}{0}=\left[ ? \right]\](i.e. *some* number)

  18. anonymous
    • 5 years ago
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    we could write

  19. anonymous
    • 5 years ago
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    \[6=0 \times \left[ ? \right]\]

  20. anonymous
    • 5 years ago
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    What number would you put in now?

  21. anonymous
    • 5 years ago
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    thats impossible

  22. anonymous
    • 5 years ago
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    EXACTLY

  23. anonymous
    • 5 years ago
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    We attempted to divide 6 by 0 and ended up with an IMPOSSIBILITY.

  24. anonymous
    • 5 years ago
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    That's why division by zero is undefined.

  25. anonymous
    • 5 years ago
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    Is that better?

  26. anonymous
    • 5 years ago
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    This will happen no matter what you try to divide by zero.

  27. anonymous
    • 5 years ago
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    yeahh

  28. anonymous
    • 5 years ago
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    So that's why you have to find all the x-values that will end up making x^2-9 = 0 ... because everything falls apart there.

  29. anonymous
    • 5 years ago
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    If x is 3 or -3, x^2-9 = 3^2-9 = 9 - 9 = 0 or x^2-9 = (-3)^2-9 = 9-9=0

  30. anonymous
    • 5 years ago
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    \[\frac{x-2}{x^2-9}=\frac{3 -2}{3^2-9}=\frac{1}{0}\]when x=3

  31. anonymous
    • 5 years ago
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    \[\frac{x-2}{x^2-9}=\frac{-3-2}{(-3)^2-9}=\frac{-5}{0}\]when x=-3.

  32. anonymous
    • 5 years ago
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    oooooooooooooh alrighttt tanks

  33. anonymous
    • 5 years ago
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    In both cases, you'll end up with the same rubbish you get with \[\frac{6}{0}\]

  34. anonymous
    • 5 years ago
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    I think it's clicked now :)

  35. anonymous
    • 5 years ago
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    Feel free to become a fan :P

  36. anonymous
    • 5 years ago
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    :D

  37. anonymous
    • 5 years ago
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    lol

  38. anonymous
    • 5 years ago
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    Good luck with your algebra. Try and link it to numbers if you have trouble.

  39. anonymous
    • 5 years ago
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    okeydokey

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