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

- anonymous

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

- schrodinger

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- precal

can't do this because complex come in pairs
3+i and 3-i are solutions

- precal

this is really a cubic

- anonymous

right this que. is wrong

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## More answers

- precal

were you asked to find a polynomial?

- precal

if so then, we can do it

- 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

If there is no requirement to have rational coefficients, then there's no problem, just multiply the factors together.

- anonymous

*(meant to say "no requirement to have 'real' coefficients, ...")

- 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

@cliffsedge hello...?

- 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

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

- 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

no i meant this: |dw:1350666676396:dw|

- anonymous

I don't know. That looks pretty weird; I'm not even sure it's quadratic.

- precal

not quadratic,
let's review the 3 possible solutions of a quadratic|dw:1350688342898:dw|one solution double root

- precal

|dw:1350688368004:dw|2solutions

- precal

|dw:1350688390750:dw|no solution 2 complex solutions
not a quadratic, no matter how you look at it

- anonymous

@precal , what if the quadratic equation has complex coefficients like the ones we found above?

- 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

The complex conjugate theorem only applies to polynomials with real coefficients @precal

- precal

quadratics are polynomials.............

- anonymous

That's right.
The complex conjugate theorem as it applies to quadratic polynomials only applies to quadratic polynomials with real coefficients.

- precal

http://www.purplemath.com/modules/polydefs.htm

- 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

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

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

- precal

QED that is only used in formal proofs, trust me you did not provide a formal proof.....still not a quadratic

- 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

http://www.icoachmath.com/math_dictionary/quadratic_function.html

- precal

http://www.purplemath.com/modules/quadform.htm

- 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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