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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
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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precal Group TitleBest ResponseYou've already chosen the best response.0
can't do this because complex come in pairs 3+i and 3i are solutions
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
this is really a cubic
 2 years ago

vipul92 Group TitleBest ResponseYou've already chosen the best response.0
right this que. is wrong
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
were you asked to find a polynomial?
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
if so then, we can do it
 2 years ago

livelaughlilz Group TitleBest ResponseYou've already chosen the best response.0
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

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
If there is no requirement to have rational coefficients, then there's no problem, just multiply the factors together.
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
*(meant to say "no requirement to have 'real' coefficients, ...")
 2 years ago

livelaughlilz Group TitleBest ResponseYou've already chosen the best response.0
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^25+2i/2x3+i/2
 2 years ago

livelaughlilz Group TitleBest ResponseYou've already chosen the best response.0
@cliffsedge hello...?
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
(Sorry, didn't see your reply) If the roots are 3+i and 1/2, the the factors are (x3i) and (x+1/2) When I multiply those together, I get \[\large x^2\frac{5}{2}xix\frac{3}{2}\frac{1}{2}i\]
 2 years ago

livelaughlilz Group TitleBest ResponseYou've already chosen the best response.0
I got that too. But the answer is supposed to be x^25+2i/2x3+i/2
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
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

livelaughlilz Group TitleBest ResponseYou've already chosen the best response.0
no i meant this: dw:1350666676396:dw
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
I don't know. That looks pretty weird; I'm not even sure it's quadratic.
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
not quadratic, let's review the 3 possible solutions of a quadraticdw:1350688342898:dwone solution double root
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1350688368004:dw2solutions
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
dw:1350688390750:dwno solution 2 complex solutions not a quadratic, no matter how you look at it
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
@precal , what if the quadratic equation has complex coefficients like the ones we found above?
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
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 3i because of the thm.....this is not a quadratic (This would be a great counterexample of a quadratic)
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
The complex conjugate theorem only applies to polynomials with real coefficients @precal
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
quadratics are polynomials.............
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
That's right. The complex conjugate theorem as it applies to quadratic polynomials only applies to quadratic polynomials with real coefficients.
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
http://www.purplemath.com/modules/polydefs.htm
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
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

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
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

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
\[\large x^2\frac{5+2i}{2}x\frac{3+i}{2}=0\] Is a quadratic equation, and it does have roots {3+i,1/2} QED.
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
QED that is only used in formal proofs, trust me you did not provide a formal proof.....still not a quadratic
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
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

precal Group TitleBest ResponseYou've already chosen the best response.0
http://www.icoachmath.com/math_dictionary/quadratic_function.html
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
http://www.purplemath.com/modules/quadform.htm
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
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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