mathslover
  • mathslover
Show that every complex number is a root of some quadratic equation.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I forgot to say : Congratulations for \(\large \color{green}{\frak{green..}}\) @mathslover
mathslover
  • mathslover
Thanks waterineyes.
klimenkov
  • klimenkov
\((x-(a+bi))(x-(a-bi))=x^2-2ax+(a^2+b^2)\) 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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More answers

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

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