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

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I forgot to say : Congratulations for \(\large \color{green}{\frak{green..}}\) @mathslover
Thanks waterineyes.
\((x-(a+bi))(x-(a-bi))=x^2-2ax+(a^2+b^2)\) 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.

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Hmn wait let me check it .. :)
What' remarkable in @klimenkov is that his coefficients are REAL
Oh k so there is a mistake. That is : Where is quadratic equation @klimenkov ? it must have been : \[\large{(X-(a+ib))(X-(a+ib))=\color{blue}{0}}\]
Ok I meant (a-ib) in one of them .. :)
ummm what subject is this?
1) Equation \(x^2-2ax+(a^2+b^2)\) is QUADRATIC. 2) Equation \(x^2-2ax+(a^2+b^2)\) has a root \(x=a+bi\). 3) \(a+bi\) is any complex numer for \(a\in \mathbb R, b \in \mathbb R\).
Try to multiply carefully, there is no mistake.
@klimenkov I was saying that, you didn't include 0 in RHS, why so?
Ah.. I'm really sorry. You are absolutely right. It was not an equation. It was just a polynomial.
Oh no problem. That was a minor mistake ... No problem.. Thanks though.
@UsukiDoll this is Mathematics...
I like questions from people who have SmartScore > 75. Ask something interesting again!
OK, if I will get a doubt, then I will like to ask you.
am i late?
But if you have no need in real coefficients, you can just make an equation \((x-(a+bi))^2=0\)
@mathslover i would just observe the graph and stat this statement is true
and would keep in mind that evry real number a can be written as a complex number a+0i
@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.
If you are asking for chapter then it is \(\mathsf{Complex \; \; Numbers...}\)
Ha ha ha ha...
And I was just kidding...
If you are asking for chapter then it is \(\text{Algebra}\).
oh my O_O

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