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

Algebra
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can't do this because complex come in pairs 3+i and 3-i are solutions
this is really a cubic
right this que. is wrong

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Other answers:

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