mathmath333
  • mathmath333
Find the range of \(x\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} x^2+x+1>0,\quad x\in \mathbb{R} \hspace{.33em}\\~\\ \end{align}}\)
ParthKohli
  • ParthKohli
\[x \in R\]
ParthKohli
  • ParthKohli
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 \(x\).

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ParthKohli
  • ParthKohli
Are you in 11th?
mathmath333
  • mathmath333
it has imaginary roots
ParthKohli
  • ParthKohli
Well, that's obvious. All polynomials have roots. It comes without saying that I mean real roots.
ParthKohli
  • ParthKohli
Anyway, do you see why that holds for all \(x\)?
mathmath333
  • mathmath333
cuz \(x\in \mathbb{R},x\cancel{\in} \mathbb{I}\)
ParthKohli
  • ParthKohli
In inequalities, we don't talk about complex numbers anyway. Do you see why \(x^2 + x + 1\) is positive for all \(x\)?
ParthKohli
  • ParthKohli
Another way to think is that the minimum value of the function is \(3/4\), so the function will always be greater than 0.
mathmath333
  • mathmath333
how u got 3/4
ParthKohli
  • ParthKohli
\[x^2 + x + 1 = \left(x + \frac{1}{2}\right) ^2 + \frac{3}{4}\]

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