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anonymous

  • one year ago

f(x) = 2x^3 - 5x^2 + px -5 = 0. I've found the roots: 1+2i and 1-2i. How do I solve the equation? Do I simply use the formula to find the quadratic? (x-(1-2i))(x-(1+2i))?

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  1. anonymous
    • one year ago
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    I found the quadratic to be: \[x ^{2}-2x+5\] What do I do from here?

  2. triciaal
    • one year ago
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    missing a root, x^3 means it will have 3 roots

  3. triciaal
    • one year ago
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    how did you find the roots with 2 unknowns?

  4. triciaal
    • one year ago
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    given the eqn factor then find what value of p makes the eqn true

  5. triciaal
    • one year ago
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    |dw:1443363394096:dw||dw:1443363662486:dw|

  6. triciaal
    • one year ago
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    |dw:1443363737329:dw|

  7. triciaal
    • one year ago
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    any questions?

  8. triciaal
    • one year ago
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    |dw:1443364057681:dw|

  9. anonymous
    • one year ago
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    I'm just reading your answer. It game me one root and the other is the complex conjugate.

  10. anonymous
    • one year ago
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    I got p = 12. (x-(1-2i))(x-(1+2i)) =x^2 -2x + 5 if you multiply that with 2x-1, you arrive at the original question. so x is also 1/2

  11. triciaal
    • one year ago
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    |dw:1443364335042:dw|

  12. triciaal
    • one year ago
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    I found the value of p the roots would then be the values of x when (2x - 5) = 0 and (x^2 + 1) = 0 x = 5/2 and I

  13. triciaal
    • one year ago
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    |dw:1443367088454:dw|

  14. triciaal
    • one year ago
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    p = 2 review your steps when you have 12 maybe it should be p+2*5 =12 maybe you forgot to multiply by the leading coefficient 2 you did not show your work

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