anonymous
  • anonymous
find a quadratic equation whose roots are 3+i and -1/2
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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precal
  • precal
can't do this because complex come in pairs 3+i and 3-i are solutions
precal
  • precal
this is really a cubic
anonymous
  • anonymous
right this que. is wrong

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precal
  • precal
were you asked to find a polynomial?
precal
  • precal
if so then, we can do it
anonymous
  • anonymous
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?
anonymous
  • anonymous
If there is no requirement to have rational coefficients, then there's no problem, just multiply the factors together.
anonymous
  • anonymous
*(meant to say "no requirement to have 'real' coefficients, ...")
anonymous
  • anonymous
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
anonymous
  • anonymous
@cliffsedge hello...?
anonymous
  • anonymous
(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 \[\large x^2-\frac{5}{2}x-ix-\frac{3}{2}-\frac{1}{2}i\]
anonymous
  • anonymous
I got that too. But the answer is supposed to be x^2-5+2i/2x-3+i/2
anonymous
  • anonymous
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.
anonymous
  • anonymous
no i meant this: |dw:1350666676396:dw|
anonymous
  • anonymous
I don't know. That looks pretty weird; I'm not even sure it's quadratic.
precal
  • precal
not quadratic, let's review the 3 possible solutions of a quadratic|dw:1350688342898:dw|one solution double root
precal
  • precal
|dw:1350688368004:dw|2solutions
precal
  • precal
|dw:1350688390750:dw|no solution 2 complex solutions not a quadratic, no matter how you look at it
anonymous
  • anonymous
@precal , what if the quadratic equation has complex coefficients like the ones we found above?
precal
  • precal
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)
anonymous
  • anonymous
The complex conjugate theorem only applies to polynomials with real coefficients @precal
precal
  • precal
quadratics are polynomials.............
anonymous
  • anonymous
That's right. The complex conjugate theorem as it applies to quadratic polynomials only applies to quadratic polynomials with real coefficients.
precal
  • precal
http://www.purplemath.com/modules/polydefs.htm
precal
  • precal
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[\large x^2-\frac{5+2i}{2}x-\frac{3+i}{2}=0\] Is a quadratic equation, and it does have roots {3+i,-1/2} QED.
precal
  • precal
QED that is only used in formal proofs, trust me you did not provide a formal proof.....still not a quadratic
anonymous
  • anonymous
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 ?
precal
  • precal
http://www.icoachmath.com/math_dictionary/quadratic_function.html
precal
  • precal
http://www.purplemath.com/modules/quadform.htm
precal
  • precal
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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