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mathslover Group Title

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

  • one year ago
  • one year ago

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  1. waterineyes Group Title
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    I forgot to say : Congratulations for \(\large \color{green}{\frak{green..}}\) @mathslover

    • one year ago
  2. mathslover Group Title
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    Thanks waterineyes.

    • one year ago
  3. klimenkov Group Title
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    \((x-(a+bi))(x-(a-bi))=x^2-2ax+(a^2+b^2)\) 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.

    • one year ago
  4. mathslover Group Title
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    Hmn wait let me check it .. :)

    • one year ago
  5. Mikael Group Title
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    What' remarkable in @klimenkov is that his coefficients are REAL

    • one year ago
  6. mathslover Group Title
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    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
  7. mathslover Group Title
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    Ok I meant (a-ib) in one of them .. :)

    • one year ago
  8. UsukiDoll Group Title
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    ummm what subject is this?

    • one year ago
  9. klimenkov Group Title
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    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 \(a\in \mathbb R, b \in \mathbb R\).

    • one year ago
  10. klimenkov Group Title
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    Try to multiply carefully, there is no mistake.

    • one year ago
  11. mathslover Group Title
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    @klimenkov I was saying that, you didn't include 0 in RHS, why so?

    • one year ago
  12. klimenkov Group Title
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    Ah.. I'm really sorry. You are absolutely right. It was not an equation. It was just a polynomial.

    • one year ago
  13. mathslover Group Title
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    Oh no problem. That was a minor mistake ... No problem.. Thanks though.

    • one year ago
  14. waterineyes Group Title
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    @UsukiDoll this is Mathematics...

    • one year ago
  15. klimenkov Group Title
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    I like questions from people who have SmartScore > 75. Ask something interesting again!

    • one year ago
  16. mathslover Group Title
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    OK, if I will get a doubt, then I will like to ask you.

    • one year ago
  17. AravindG Group Title
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    am i late?

    • one year ago
  18. klimenkov Group Title
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    But if you have no need in real coefficients, you can just make an equation \((x-(a+bi))^2=0\)

    • one year ago
  19. AravindG Group Title
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    @mathslover i would just observe the graph and stat this statement is true

    • one year ago
  20. AravindG Group Title
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    and would keep in mind that evry real number a can be written as a complex number a+0i

    • one year ago
  21. UsukiDoll Group Title
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    @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
  22. waterineyes Group Title
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    If you are asking for chapter then it is \(\mathsf{Complex \; \; Numbers...}\)

    • one year ago
  23. UsukiDoll Group Title
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    o_O

    • one year ago
  24. waterineyes Group Title
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    Ha ha ha ha...

    • one year ago
  25. waterineyes Group Title
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    And I was just kidding...

    • one year ago
  26. klimenkov Group Title
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    If you are asking for chapter then it is \(\text{Algebra}\).

    • one year ago
  27. UsukiDoll Group Title
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    oh my O_O

    • one year ago
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