A community for students.
Here's the question you clicked on:
 0 viewing
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.

This Question is Closed

waterineyes
 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 .. :)

Mikael
 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.

waterineyes
 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.

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

waterineyes
 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 \(\text{Algebra}\).
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.