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kitsune0724 Group Title

By factoring how do you solve 3x^2/3-11x^1/3-4=0? My teacher doesn't do substitution. please and thank you.

  • 2 years ago
  • 2 years ago

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  1. cwrw238 Group Title
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    (3x^1/3 + 1)(x^1/3 - 4 ) = 0

    • 2 years ago
  2. cwrw238 Group Title
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    can you continue?

    • 2 years ago
  3. cwrw238 Group Title
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    3x^1/3 + 1 = 0 x^(1/3) = -1/3 x^1/3 - 4 = 0 x^1/3 = 4

    • 2 years ago
  4. kitsune0724 Group Title
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    let me see...

    • 2 years ago
  5. cwrw238 Group Title
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    hint: x = x(1/3)^3

    • 2 years ago
  6. kitsune0724 Group Title
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    (-1/3)^3 or (1/3)^3?

    • 2 years ago
  7. cwrw238 Group Title
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    sorry - i didn't write it corruptly (x^(1/3)^ 3 = x^(1/3 * 3) = x^1 = x or in other words you have the cube root of and you cube it - that gives you x so if you cube both sides of the equations you get the value of x e g (x^(1/3)^ 3 =( -1/3)^3 so x = - 1/3^3 = -1/27 thats one of the roots

    • 2 years ago
  8. cwrw238 Group Title
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    * correctly

    • 2 years ago
  9. kitsune0724 Group Title
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    Thank you so much.

    • 2 years ago
  10. cwrw238 Group Title
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    note x^(1/3) is same as cube root x

    • 2 years ago
  11. cwrw238 Group Title
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    so to find other root you do same thing with x^1/3 = 4

    • 2 years ago
  12. cwrw238 Group Title
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    yw

    • 2 years ago
  13. kitsune0724 Group Title
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    Thank you again. But do you divide x^1/3=4? Like 4/1=4.

    • 2 years ago
  14. cwrw238 Group Title
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    just cube both sides of th equation as we did to the first one (x^(1/3)^3 = x 4^3 = 4*4*4 = 64 x = 64

    • 2 years ago
  15. cwrw238 Group Title
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    i suggest you review the laws of indices x^a + x^b = x^(a+b) x^a / x^b = x^(a-b) (x^a)^b = x^(a*b) = x^ ab

    • 2 years ago
  16. cwrw238 Group Title
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    also |dw:1342632977864:dw|

    • 2 years ago
  17. cwrw238 Group Title
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    please ask if you need any more help

    • 2 years ago
  18. kitsune0724 Group Title
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    Thank you so much.

    • 2 years ago
  19. cwrw238 Group Title
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    yw

    • 2 years ago
  20. kitsune0724 Group Title
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    Could you please help with another problem?

    • 2 years ago
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