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SheldonEinstein

  • 3 years ago

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

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

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

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

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

  5. ParthKohli
    • 3 years ago
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    Hindi mein bolun?

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

  7. SheldonEinstein
    • 3 years ago
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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 = \(\LARGE\frac{-2 \pm \sqrt{4-12}}{2} = \frac{-2 \pm 2\sqrt{-2}}{2} = -1 \pm \sqrt{2} i \)

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

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

  10. SheldonEinstein
    • 3 years ago
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    OK thanks for the shortcut friend...

  11. ParthKohli
    • 3 years ago
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    |z| is the modulus?

  12. SheldonEinstein
    • 3 years ago
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    Yeah

  13. SheldonEinstein
    • 3 years ago
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    It is modulus.

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

  15. SheldonEinstein
    • 3 years ago
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    :) Yeah Yeah Yeah!!! Thanks!

  16. ParthKohli
    • 3 years ago
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    You're welcome ;)

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