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SheldonEinstein Group Title

The polynomial \(P(X) = a_0 + a_1X + a_2X^2 + ....+ a_nX^n\) where \(a_0,a_1,...,a_n\) are all real numbers and \(a_0>a_1>a_2>....>a_n>0\). Then which of the following MUST be true: P(X) always has a real root P(X) has a complex root z and |z| ≥1 P(X) has a complex root and |z|<1 P(X) does not have any complex root

  • 2 years ago
  • 2 years ago

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  1. ParthKohli Group Title
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    Always assume a polynomial of this form. Use this for the experiments:\[\rm p(x) = x^2 + 2x + 3\]

    • 2 years ago
  2. ParthKohli Group Title
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    Now see what all is true.

    • 2 years ago
  3. SheldonEinstein Group Title
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    @ParthKohli Sorry, didn't get you, can you explain me more?

    • 2 years ago
  4. ParthKohli Group Title
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    In these questions, I always take an example of a polynomial and see what all is true amongst the choices.

    • 2 years ago
  5. ParthKohli Group Title
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    Hindi mein bolun?

    • 2 years ago
  6. SheldonEinstein Group Title
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    OK... i am getting it... let me check the choices now.

    • 2 years ago
  7. SheldonEinstein Group Title
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    At least it will be complex root if I take your equation of the polynomial. P(X) = \(x^2+2x+3\) x = \(\LARGE\frac{-2 \pm \sqrt{4-12}}{2} = \frac{-2 \pm 2\sqrt{-2}}{2} = -1 \pm \sqrt{2} i \)

    • 2 years ago
  8. SheldonEinstein Group Title
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    So it will be ,,,, 2nd option : P(X) has a complex root z and |z| ≥1 Right?

    • 2 years ago
  9. ParthKohli Group Title
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    Keep taking such equations and you'd get your answer... I think.

    • 2 years ago
  10. SheldonEinstein Group Title
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    OK thanks for the shortcut friend...

    • 2 years ago
  11. ParthKohli Group Title
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    |z| is the modulus?

    • 2 years ago
  12. SheldonEinstein Group Title
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    Yeah

    • 2 years ago
  13. SheldonEinstein Group Title
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    It is modulus.

    • 2 years ago
  14. ParthKohli Group Title
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    \[\sqrt{(-1)^2 + (\sqrt2)^2} = \sqrt{1 + 2} = \sqrt 3\]Yes, you're right

    • 2 years ago
  15. SheldonEinstein Group Title
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    :) Yeah Yeah Yeah!!! Thanks!

    • 2 years ago
  16. ParthKohli Group Title
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    You're welcome ;)

    • 2 years ago
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