anonymous
  • anonymous
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
Mathematics
schrodinger
  • schrodinger
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ParthKohli
  • ParthKohli
Always assume a polynomial of this form. Use this for the experiments:\[\rm p(x) = x^2 + 2x + 3\]
ParthKohli
  • ParthKohli
Now see what all is true.
anonymous
  • anonymous
@ParthKohli Sorry, didn't get you, can you explain me more?

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ParthKohli
  • ParthKohli
In these questions, I always take an example of a polynomial and see what all is true amongst the choices.
ParthKohli
  • ParthKohli
Hindi mein bolun?
anonymous
  • anonymous
OK... i am getting it... let me check the choices now.
anonymous
  • anonymous
At least it will be complex root if I take your equation of the polynomial. P(X) = \(x^2+2x+3\) x = \(\LARGE\frac{-2 \pm \sqrt{4-12}}{2} = \frac{-2 \pm 2\sqrt{-2}}{2} = -1 \pm \sqrt{2} i \)
anonymous
  • anonymous
So it will be ,,,, 2nd option : P(X) has a complex root z and |z| ≥1 Right?
ParthKohli
  • ParthKohli
Keep taking such equations and you'd get your answer... I think.
anonymous
  • anonymous
OK thanks for the shortcut friend...
ParthKohli
  • ParthKohli
|z| is the modulus?
anonymous
  • anonymous
Yeah
anonymous
  • anonymous
It is modulus.
ParthKohli
  • ParthKohli
\[\sqrt{(-1)^2 + (\sqrt2)^2} = \sqrt{1 + 2} = \sqrt 3\]Yes, you're right
anonymous
  • anonymous
:) Yeah Yeah Yeah!!! Thanks!
ParthKohli
  • ParthKohli
You're welcome ;)

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