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livelaughlilz Group Title

find a quadratic equation whose roots are 3+i and -1/2

  • 2 years ago
  • 2 years ago

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  1. precal Group Title
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    can't do this because complex come in pairs 3+i and 3-i are solutions

    • 2 years ago
  2. precal Group Title
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    this is really a cubic

    • 2 years ago
  3. vipul92 Group Title
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    right this que. is wrong

    • 2 years ago
  4. precal Group Title
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    were you asked to find a polynomial?

    • 2 years ago
  5. precal Group Title
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    if so then, we can do it

    • 2 years ago
  6. livelaughlilz Group Title
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    no i was asked to find a quadratic. wouldn't the 3 + i and 3 - i pair only be required if i was asked for integral coefficients?

    • 2 years ago
  7. CliffSedge Group Title
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    If there is no requirement to have rational coefficients, then there's no problem, just multiply the factors together.

    • 2 years ago
  8. CliffSedge Group Title
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    *(meant to say "no requirement to have 'real' coefficients, ...")

    • 2 years ago
  9. livelaughlilz Group Title
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    i know the formula x^2 - (sum of the roots)x + product of the roots = 0. but i did it and got x^2 - (5/2 + i)x - 2i = 0. the answer is supposed to be x^2-5+2i/2x-3+i/2

    • 2 years ago
  10. livelaughlilz Group Title
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    @cliffsedge hello...?

    • 2 years ago
  11. CliffSedge Group Title
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    (Sorry, didn't see your reply) If the roots are 3+i and -1/2, the the factors are (x-3-i) and (x+1/2) When I multiply those together, I get \[\large x^2-\frac{5}{2}x-ix-\frac{3}{2}-\frac{1}{2}i\]

    • 2 years ago
  12. livelaughlilz Group Title
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    I got that too. But the answer is supposed to be x^2-5+2i/2x-3+i/2

    • 2 years ago
  13. CliffSedge Group Title
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    I'm not sure I understand that answer. You mean this? \[\large x^2-\frac{(5+2i)}{2}x-\frac{(3+i)}{2}\] If so, then they are equivalent, try simplifying the above.

    • 2 years ago
  14. livelaughlilz Group Title
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    no i meant this: |dw:1350666676396:dw|

    • 2 years ago
  15. CliffSedge Group Title
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    I don't know. That looks pretty weird; I'm not even sure it's quadratic.

    • 2 years ago
  16. precal Group Title
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    not quadratic, let's review the 3 possible solutions of a quadratic|dw:1350688342898:dw|one solution double root

    • 2 years ago
  17. precal Group Title
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    |dw:1350688368004:dw|2solutions

    • 2 years ago
  18. precal Group Title
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    |dw:1350688390750:dw|no solution 2 complex solutions not a quadratic, no matter how you look at it

    • 2 years ago
  19. CliffSedge Group Title
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    @precal , what if the quadratic equation has complex coefficients like the ones we found above?

    • 2 years ago
  20. precal Group Title
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    not possible, complex conjugate theorem states that complex solutions come in pairs. Even if you are not given both, you can assume the other by the thm if 3+i is a known zero, we can assume 3-i because of the thm.....this is not a quadratic (This would be a great counterexample of a quadratic)

    • 2 years ago
  21. CliffSedge Group Title
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    The complex conjugate theorem only applies to polynomials with real coefficients @precal

    • 2 years ago
  22. precal Group Title
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    quadratics are polynomials.............

    • 2 years ago
  23. CliffSedge Group Title
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    That's right. The complex conjugate theorem as it applies to quadratic polynomials only applies to quadratic polynomials with real coefficients.

    • 2 years ago
  24. precal Group Title
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    http://www.purplemath.com/modules/polydefs.htm

    • 2 years ago
  25. precal Group Title
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    I think you are misinterpreting the thm..........either way it is not a quadratic I suspect it is a cubic then it would make sense

    • 2 years ago
  26. CliffSedge Group Title
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    Nope. The theorem states that *if* a polynomial has real coefficients then complex roots will come in conjugate pairs. If the polynomials does not have real coefficients then the theorem does not apply.

    • 2 years ago
  27. CliffSedge Group Title
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    \[\large x^2-\frac{5+2i}{2}x-\frac{3+i}{2}=0\] Is a quadratic equation, and it does have roots {3+i,-1/2} QED.

    • 2 years ago
  28. precal Group Title
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    QED that is only used in formal proofs, trust me you did not provide a formal proof.....still not a quadratic

    • 2 years ago
  29. CliffSedge Group Title
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    I was just being cute with the "QED" And yes it is a quadratic equation. Can you provide any sources to back up your claim that it is not @precal ?

    • 2 years ago
  30. precal Group Title
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    http://www.icoachmath.com/math_dictionary/quadratic_function.html

    • 2 years ago
  31. precal Group Title
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    http://www.purplemath.com/modules/quadform.htm

    • 2 years ago
  32. precal Group Title
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    I could go on but then that would be me being "cute" bottom line is complex solutions come in pairs..............If you know one, you can assume the other.....

    • 2 years ago
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