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Mimi_x3

The polynomial P(x) = x^3 - 2x^2 + kx + 24 has roots a , b and γ (i) It is known that two of the roots are equal in magnitude but opposite in sign. Find the third root and hence find the value of k Can you please help .

  • 2 years ago
  • 2 years ago

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  1. oktalBlizzard
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    :@

    • 2 years ago
  2. Mimi_x3
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    :@ ?? why yu angry ? LOL

    • 2 years ago
  3. aredsani
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    3rd root is 2

    • 2 years ago
  4. oktalBlizzard
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    not angry

    • 2 years ago
  5. oktalBlizzard
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    just having a bad day

    • 2 years ago
  6. Mimi_x3
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    how is 3rd root 2 ?

    • 2 years ago
  7. Mimi_x3
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    lols , having a bad day ? why ?

    • 2 years ago
  8. oktalBlizzard
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    I passed the entry test to uni, but apparently, I need to get minimum of 2 A levels to get into it. A levels take 2 years to get. I have 4 months.

    • 2 years ago
  9. Mimi_x3
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    ohh ok LOL

    • 2 years ago
  10. estudier
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    I got so far (for roots a,-a and c) 3a +c = 2 and a^2 c = 24 U get that?

    • 2 years ago
  11. Mimi_x3
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    yeah

    • 2 years ago
  12. Mimi_x3
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    w8 , what's c ?

    • 2 years ago
  13. estudier
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    I don't mean understand, I mean u calculated same?

    • 2 years ago
  14. estudier
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    I said for roots, a,-a and c

    • 2 years ago
  15. Mimi_x3
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    ohh lols , i dnt even know how to do it

    • 2 years ago
  16. estudier
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    There is similar (not same) formula like Vieta for cubic as for quadratic....

    • 2 years ago
  17. Mimi_x3
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    What's Vieta ?

    • 2 years ago
  18. estudier
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    Formulae for sum and product of roots of quadratic.

    • 2 years ago
  19. Mimi_x3
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    ohh that

    • 2 years ago
  20. jimmywhoaaw
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    Let a be the first root >0. Then b=-a Now since a cubic function with 3 real roots then must be of the form: (x-a)(x-b)(x-v)= x^3 - 2x^2 + kx + 24 But b=-a so LHS is (x-a)(x+a)(x-v)=(x^2-a^2)(x-v) Multiplying the left hand side out we get: =x^3-v*x^2-a^2*x+v*a^2=P(x) From this we see that v=2 which implies 24=2*a^2 then a^2=12 but -a^2*x=kx thus k=-12

    • 2 years ago
  21. Mimi_x3
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    thank you (:

    • 2 years ago
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