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Show that every complex number is a root of some quadratic equation.
 one year ago
 one year ago
Show that every complex number is a root of some quadratic equation.
 one year ago
 one year ago

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

mathsloverBest ResponseYou've already chosen the best response.0
Thanks waterineyes.
 one year ago

klimenkovBest 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.
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
Hmn wait let me check it .. :)
 one year ago

MikaelBest ResponseYou've already chosen the best response.0
What' remarkable in @klimenkov is that his coefficients are REAL
 one year ago

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

mathsloverBest ResponseYou've already chosen the best response.0
Ok I meant (aib) in one of them .. :)
 one year ago

UsukiDollBest ResponseYou've already chosen the best response.0
ummm what subject is this?
 one year ago

klimenkovBest ResponseYou've already chosen the best response.2
1) 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 \(a\in \mathbb R, b \in \mathbb R\).
 one year ago

klimenkovBest ResponseYou've already chosen the best response.2
Try to multiply carefully, there is no mistake.
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
@klimenkov I was saying that, you didn't include 0 in RHS, why so?
 one year ago

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

mathsloverBest ResponseYou've already chosen the best response.0
Oh no problem. That was a minor mistake ... No problem.. Thanks though.
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
@UsukiDoll this is Mathematics...
 one year ago

klimenkovBest ResponseYou've already chosen the best response.2
I like questions from people who have SmartScore > 75. Ask something interesting again!
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
OK, if I will get a doubt, then I will like to ask you.
 one year ago

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

AravindGBest ResponseYou've already chosen the best response.1
@mathslover i would just observe the graph and stat this statement is true
 one year ago

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

UsukiDollBest 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.
 one year ago

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

waterineyesBest ResponseYou've already chosen the best response.0
And I was just kidding...
 one year ago

klimenkovBest ResponseYou've already chosen the best response.2
If you are asking for chapter then it is \(\text{Algebra}\).
 one year ago
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