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mathmath333

  • one year ago

Find the range of \(x\)

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} x^2+x+1>0,\quad x\in \mathbb{R} \hspace{.33em}\\~\\ \end{align}}\)

  2. ParthKohli
    • one year ago
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    \[x \in R\]

  3. ParthKohli
    • one year ago
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    This quadratic expression has no roots, and it points upwards. It means that it must lie above the x-axis always, thus meaning that the inequality holds for all \(x\).

  4. ParthKohli
    • one year ago
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    Are you in 11th?

  5. mathmath333
    • one year ago
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    it has imaginary roots

  6. ParthKohli
    • one year ago
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    Well, that's obvious. All polynomials have roots. It comes without saying that I mean real roots.

  7. ParthKohli
    • one year ago
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    Anyway, do you see why that holds for all \(x\)?

  8. mathmath333
    • one year ago
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    cuz \(x\in \mathbb{R},x\cancel{\in} \mathbb{I}\)

  9. ParthKohli
    • one year ago
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    In inequalities, we don't talk about complex numbers anyway. Do you see why \(x^2 + x + 1\) is positive for all \(x\)?

  10. ParthKohli
    • one year ago
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    Another way to think is that the minimum value of the function is \(3/4\), so the function will always be greater than 0.

  11. mathmath333
    • one year ago
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    how u got 3/4

  12. ParthKohli
    • one year ago
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    \[x^2 + x + 1 = \left(x + \frac{1}{2}\right) ^2 + \frac{3}{4}\]

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