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mathslover
 3 years ago
Show that every complex number is a root of some quadratic equation.
mathslover
 3 years ago
Show that every complex number is a root of some quadratic equation.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I forgot to say : Congratulations for \(\large \color{green}{\frak{green..}}\) @mathslover

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2\((x(a+bi))(x(abi))=x^22ax+(a^2+b^2)\) So, we get that this quadratic equation has a root \(x=a+bi\). So as \(a\) and \(b\) are any real numbers we prove it for any complex number.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Hmn wait let me check it .. :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What' remarkable in @klimenkov is that his coefficients are REAL

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Oh k so there is a mistake. That is : Where is quadratic equation @klimenkov ? it must have been : \[\large{(X(a+ib))(X(a+ib))=\color{blue}{0}}\]

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Ok I meant (aib) in one of them .. :)

UsukiDoll
 3 years ago
Best ResponseYou've already chosen the best response.0ummm what subject is this?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.21) Equation \(x^22ax+(a^2+b^2)\) is QUADRATIC. 2) Equation \(x^22ax+(a^2+b^2)\) has a root \(x=a+bi\). 3) \(a+bi\) is any complex numer for \(a\in \mathbb R, b \in \mathbb R\).

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2Try to multiply carefully, there is no mistake.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0@klimenkov I was saying that, you didn't include 0 in RHS, why so?

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2Ah.. I'm really sorry. You are absolutely right. It was not an equation. It was just a polynomial.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Oh no problem. That was a minor mistake ... No problem.. Thanks though.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@UsukiDoll this is Mathematics...

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2I like questions from people who have SmartScore > 75. Ask something interesting again!

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0OK, if I will get a doubt, then I will like to ask you.

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2But if you have no need in real coefficients, you can just make an equation \((x(a+bi))^2=0\)

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1@mathslover i would just observe the graph and stat this statement is true

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.1and would keep in mind that evry real number a can be written as a complex number a+0i

UsukiDoll
 3 years ago
Best ResponseYou've already chosen the best response.0@wwaterineyes besides Mathematics. I meant what level of Mathematics is this because this certainly doesn't look like calculus or any lower level math question.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you are asking for chapter then it is \(\mathsf{Complex \; \; Numbers...}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And I was just kidding...

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2If you are asking for chapter then it is \(\text{Algebra}\).
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