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

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

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

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

livelaughlilz
 2 years ago
Best ResponseYou've already chosen the best response.0no 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?

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

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

livelaughlilz
 2 years ago
Best ResponseYou've already chosen the best response.0i 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

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

CliffSedge
 2 years ago
Best 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 \[\large x^2\frac{5}{2}xix\frac{3}{2}\frac{1}{2}i\]

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

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.1I'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.

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

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

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

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

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

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

precal
 2 years ago
Best ResponseYou've already chosen the best response.0not 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)

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

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

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

precal
 2 years ago
Best ResponseYou've already chosen the best response.0I think you are misinterpreting the thm..........either way it is not a quadratic I suspect it is a cubic then it would make sense

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.1Nope. 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.

CliffSedge
 2 years ago
Best 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.

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

CliffSedge
 2 years ago
Best ResponseYou've already chosen the best response.1I 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
 2 years ago
Best ResponseYou've already chosen the best response.0http://www.icoachmath.com/math_dictionary/quadratic_function.html

precal
 2 years ago
Best ResponseYou've already chosen the best response.0I 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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