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livelaughlilz

  • 2 years ago

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

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  1. precal
    • 2 years ago
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    can't do this because complex come in pairs 3+i and 3-i are solutions

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

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

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

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

  6. livelaughlilz
    • 2 years ago
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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?

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

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

  9. livelaughlilz
    • 2 years ago
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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

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

  11. CliffSedge
    • 2 years ago
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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 \[\large x^2-\frac{5}{2}x-ix-\frac{3}{2}-\frac{1}{2}i\]

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

  13. CliffSedge
    • 2 years ago
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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.

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

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

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

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

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

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

  20. precal
    • 2 years ago
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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)

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

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

  23. CliffSedge
    • 2 years ago
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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.

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

  25. precal
    • 2 years ago
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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

  26. CliffSedge
    • 2 years ago
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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.

  27. CliffSedge
    • 2 years ago
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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.

  28. precal
    • 2 years ago
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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

  29. CliffSedge
    • 2 years ago
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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 ?

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

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

  32. precal
    • 2 years ago
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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.....

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