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mazehaq

  • 2 years ago

3(x-3)^1/3 - 4(x-3)^4/3 + (x-3)^7/3 = 0 To make it easier, I substituted (x-3) with x, in which I got x^1/3(3 - 4x +x^2) ...but it can't be factored. Help me out! It's a fairly easy question I bet, but I need a nice way of solving it. Thank you! :)

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  1. lgbasallote
    • 2 years ago
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    actually you might want to substitute x - 3 with a instead to avoid confusion

  2. lgbasallote
    • 2 years ago
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    but you're right anyway... except for the last part

  3. lgbasallote
    • 2 years ago
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    wait on second thought you're right

  4. lgbasallote
    • 2 years ago
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    and it can be factored... \[\huge a^{\frac 13} (3 - 4a + a^2) \implies a^{\frac 13} (a^2 - 4a + 3)\] can you see it now?

  5. lgbasallote
    • 2 years ago
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    a^2 - 4a + 3 is factorable

  6. mazehaq
    • 2 years ago
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    Oh okay! I kept thinking that it was a^2 + 4a - 3

  7. lgbasallote
    • 2 years ago
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    oh. no wonder

  8. lgbasallote
    • 2 years ago
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    yeah. that one isn't factorable

  9. mazehaq
    • 2 years ago
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    so 3 - 4a + a^2 is a^2 - 4a + 3 ...right?

  10. lgbasallote
    • 2 years ago
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    yes

  11. mazehaq
    • 2 years ago
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    Okie dokie thank you!

  12. lgbasallote
    • 2 years ago
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    welcome

  13. mazehaq
    • 2 years ago
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    So wait, how would I figure out the final answer, I am at (x-3)^1/3 (x-7)(x-7)=0

  14. lgbasallote
    • 2 years ago
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    they shouldn't be equal...

  15. lgbasallote
    • 2 years ago
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    how did you factor a^2 - 4a + 3?

  16. mazehaq
    • 2 years ago
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    I skipped a few steps. Remember a = (x-3) a^1/3 (a-1)(a-3) = 0 (x-3)^1/3 [(x-3-1)(x-3-3)] = 0 (x-3)^1/3 (x-4)(x-6) = 0

  17. mazehaq
    • 2 years ago
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    sorry about my previous comment tho, I totally typed it wrong, my computer is lagging so bad.

  18. lgbasallote
    • 2 years ago
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    right. that's better

  19. mazehaq
    • 2 years ago
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    So how do I solve the end?

  20. mazehaq
    • 2 years ago
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    (x-3)^1/3 (x-4)(x-6) = 0

  21. lgbasallote
    • 2 years ago
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    that s the end

  22. lgbasallote
    • 2 years ago
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    isn't it?

  23. lgbasallote
    • 2 years ago
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    or do you want x?

  24. mazehaq
    • 2 years ago
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    don't I make each = to zero

  25. mazehaq
    • 2 years ago
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    Like, 2 of my answers will be 4 & 6...

  26. lgbasallote
    • 2 years ago
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    right

  27. lgbasallote
    • 2 years ago
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    now..just equate (x-3)^1/3 to 0

  28. mazehaq
    • 2 years ago
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    I know but how :(

  29. lgbasallote
    • 2 years ago
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    \[(x - 3)^{\frac 13} = 0\]

  30. lgbasallote
    • 2 years ago
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    take the cube root of both sides..what happens?

  31. mazehaq
    • 2 years ago
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    x-3=0?

  32. lgbasallote
    • 2 years ago
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    yes

  33. mazehaq
    • 2 years ago
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    therefore x = 3?

  34. lgbasallote
    • 2 years ago
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    now solve for x

  35. lgbasallote
    • 2 years ago
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    right

  36. lgbasallote
    • 2 years ago
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    so there's your answer

  37. mazehaq
    • 2 years ago
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    Okay, so answer is 3,4,6 :) thanks!

  38. lgbasallote
    • 2 years ago
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    yup and welcome

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